On the Sequential Joint (f,δ,q)-Numerical Radius

Abstract

In this paper, a new sequential joint $(f,\delta ,q)$-numerical radius is introduced for sequences of bounded linear operators on Hilbert spaces, motivated by recent generalizations of the numerical radius. Firstly, several fundamental properties of the sequential joint $(f,\delta ,q)$-numerical radius function are established. Subsequently, lower and upper bounds for the sequential joint $(f,\delta ,q)$-numerical radius function are derived, yielding generalizations of several known inequalities. Finally, the sequential joint $(f,\delta ,q)$-numerical radius function is investigated for operator sequences whose coordinate operators are sectorial, leading to further estimates and properties. The obtained results extend and unify a variety of existing results on numerical radius inequalities and provide new contributions to the theory of operator sequences on Hilbert spaces.

1. Introduction

The numerical range and the numerical radius of an operator have been investigated extensively in operator theory and have been related to various problems arising in functional analysis and its applications. In particular, several properties and inequalities involving the numerical radius have been established in connection with operator theory, numerical analysis, and related areas [1,2,3]. The investigation of numerical radius inequalities and their refinements has attracted considerable attention in the theory of operators on Hilbert spaces.

Throughout this paper, a complex Hilbert space will be denoted by H, with inner product $⟨·,·⟩$ and associated norm $\parallel ·\parallel$. The algebra of all bounded linear operators on H will be denoted by $B\left(H\right)$, and the unit sphere of H will be denoted by $S^1\left(H\right)$.

For an operator $A\in B\left(H\right)$, the numerical radius is given by

$$\omega \left(A\right)=sup\left\{\right|⟨Ax,x⟩|:x\in S^1\left(H\right)\}.$$

The operator norm of $A\in B\left(H\right)$ is given by

$$\parallel A\parallel =sup\{\parallel Ax\parallel :x\in S^1\left(H\right)\}.$$

For a general operator $A\in B\left(H\right)$, the exact value of the numerical radius $\omega \left(A\right)$ is usually difficult to determine.

Closed-form expressions can generally be obtained only when additional structural properties of the operator are available.

For this reason, considerable attention has been given to establishing upper and lower estimates for $\omega \left(A\right)$.

The numerical radius $\omega (·)$ is known to define a norm on $B\left(H\right)$. A fundamental pair of estimates for $\omega \left(A\right)$ is given by the following two-sided inequality (see, e.g., [4]):

$${\displaystyle \frac{1}{2}}\parallel A\parallel \le \omega \left(A\right)\le \parallel A\parallel .$$

(1)

Equality in the upper estimate in (1) is attained whenever A is normal, whereas equality in the lower estimate is obtained if $A^2=0$, where 0 denotes the zero operator in $B\left(H\right)$.

Several refinements of (1) have been established in the literature. In particular, the following two-sided estimate will be used here; see [5,6]:

$$\omega \left(A\right)\le {\displaystyle \frac{1}{2}}∥\left|A\right|+|A^{*}|∥\le {\displaystyle \frac{1}{2}}\left(\parallel A\parallel +\sqrt{\parallel A^2\parallel }\right),$$

(2)

$${\displaystyle \frac{1}{4}}∥A^{*}A+A{A}^{*}∥\le {\omega }^2\left(A\right)\le {\displaystyle \frac{1}{2}}∥A^{*}A+A{A}^{*}∥.$$

(3)

The inequality (3) refines both the left and right inequalities of (1).

Besides the classical numerical radius, the $\delta$-numerical range and the associated $\delta$-numerical radius provide an extension of this concept. The corresponding definition is recalled below.

Definition 1 

([7]). Let $A\in B\left(H\right)$ and let $0\le \delta \le \parallel A\parallel$. The δ-numerical range associated with A is given by

$$W_{\delta }\left(A\right)=cl\left(\left\{⟨Ax,x⟩:x\in S^1\left(H\right),\parallel Ax\parallel \ge \delta \right\}\right),$$

where $cl(·)$ stands for the closure of the corresponding set in the complex plane.

The set $W_{\delta }\left(A\right)$ is a closed subset of the closure of the classical numerical range. By applying a slight modification of a theorem due to Dekker [8], its connectedness can be established. The convexity of $W_{\delta }\left(A\right)$ is also of particular interest; in particular, convexity is ensured when A is normal or when H has dimension two.

Definition 2. 

Let $A\in B\left(H\right)$ with $0\le \delta \le \parallel A\parallel$. The quantities ${\omega }_{\delta }\left(A\right)$ and ${\parallel A\parallel }_{\delta }$ are introduced, respectively, by

$${\omega }_{\delta }\left(A\right)=sup\left\{\right|\lambda |:\lambda \in W_{\delta }\left(A\right)\}$$

and

$${\parallel A\parallel }_{\delta }=sup\left\{\parallel Ax\parallel :x\in S^1\left(H\right),\parallel Ax\parallel \ge \delta \right\}.$$

The $\delta$-numerical radius was introduced by Cho and Takaguchi [9] based on the earlier ideas of Stampfli [7].

Several estimates for the numerical radius have been obtained in terms of the norm of the operator, its adjoint, its modulus, and combinations of these quantities. Further details can be found in [4,10,11,12,13,14,15,16,17,18,19,20,21].

Another important extension of the classical numerical range and the classical numerical radius is provided by the concept of the q-numerical range and the corresponding q-numerical radius.

Definition 3 

([22]). Let $A\in B\left(H\right)$ and $q\in \mathbb{C}$ satisfy $\left|q\right|\le 1.$ The q-numerical range $W_q\left(A\right)$ and the q-numerical radius ${\omega }_q\left(A\right)$ of A are defined, respectively, as

$$W_q\left(A\right)=\{⟨Ax,y⟩:(x,y)\in S_q^1\left(H\right)\}$$

and

$${\omega }_q\left(A\right)=\underset{z\in W_q\left(A\right)}{sup}\left|z\right|,$$

where $S_q^1\left(H\right)=\left\{(x,y)\in H\times H:\parallel x\parallel =\parallel y\parallel =1,⟨x,y⟩=q\right\}.$

For $q=1$, it is readily seen that $W_q\left(A\right)$ coincides with the classical numerical range $W\left(A\right)$, and hence ${\omega }_1(·)=\omega (·)$. The q-numerical range was first introduced by Marcus and Andersen [23] in 1977 for a linear transformation A defined on an n-dimensional unitary space. The convexity of the q-numerical range was established by Tsing [24]. Various properties of $W_q\left(A\right)$ were subsequently studied by Li et al. [25] and Li and Nakazato [26]. Chien and Nakazato [27] described the boundary of a square matrix’s q-numerical range by employing the concept of the Davis–Wielandt shell. The q-numerical ranges associated with shift operators were also investigated in [28,29]. Duan [30] highlighted the significance of the q-numerical range in characterizing the perfect distinguishability of quantum operations. More recently, Moghaddam et al. [31] investigated several bounds for the q-numerical radius. Some of the bounds obtained in that work are listed below:

$$\begin{array}{c}{\displaystyle \frac{q}{2(2-q^2)}}\parallel A\parallel \le {\omega }_q\left(A\right)\le \parallel A\parallel ,\hfill \\ {\omega }_q^2\left(A\right)\le {\displaystyle \frac{q^2}{4}}{\left(\parallel A\parallel +\sqrt{\parallel A^2\parallel }\right)}^2+(1-q^2+2q\sqrt{1-q^2}){\parallel A\parallel }^2,\hfill \\ {\displaystyle \frac{q^2}{4{(2-q^2)}^2}}\parallel A^{*}A+A{A}^{*}\parallel \le {\omega }_q^2\left(A\right)\le {\displaystyle \frac{q^2}{2{(1-\sqrt{1-q^2})}^2}}\parallel A^{*}A+A{A}^{*}\parallel .\hfill \end{array}$$

These inequalities generalize the corresponding inequalities in (1), (2), and (3), respectively, for the classical numerical radius.

A sharper lower bound for the q-numerical radius was obtained in [22], namely,

$${\displaystyle \frac{q}{2}}\parallel A\parallel \le {\omega }_q\left(A\right)\le \parallel A\parallel .$$

Another important class of operators that will be used later in this paper is one special class of sectorial operators.

Definition 4. 

For any fixed $0\le \alpha \le {\displaystyle \frac{\pi }{2}},$ let

$$S\left(\alpha \right):=\{r{e}^{i\phi }:r\ge 0,-\alpha \le \phi \le \alpha \}\subset \mathbb{C}.$$

1.

An operator $A:H\to H$, with $A\in B\left(H\right)$, is called $(q,\alpha )$-sectorial on a Hilbert space H whenever

$$W_q\left(A\right)\subset S\left(\alpha \right).$$

2.

The notation $Se{c}_{(q,\alpha )}\left(H\right)$ will be used for the class of all $(q,\alpha )$-sectorial operators on the Hilbert space H, with vertex at the origin and half-angle $\alpha \in \left[0,{\displaystyle \frac{\pi }{2}}\right)$.

Note that each operator $A\in B\left(H\right)$ can be expressed in the Cartesian decomposition form as $A=ReA+iImA,$ where $ReA={\displaystyle \frac{A+A^{*}}{2}}$ and $ImA={\displaystyle \frac{A-A^{*}}{2i}}.$ Here, $A^{*}$ denotes the Hilbert adjoint of $A.$ Throughout this study we denote by $\left|A\right|={\left(A^{*}A\right)}^{1/2}$ the absolute value of an operator $A\in B\left(H\right).$ Also, we define

$$B^n\left(H\right):=B\left(H\right)\times \cdots \times B\left(H\right)\left(n\mathrm{times}\right),n\ge 1.$$

Motivated by these developments and the recent study of joint $(f,\delta )$-numerical radius functions in [32], in this paper we introduce a new sequential joint $(f,\delta ,q)$-numerical radius for operator sequences on Hilbert spaces. The choice of the parameters f, $\delta$, and q is motivated by three complementary directions in the existing generalizations of the numerical radius. The function f provides a flexible scalarizing mechanism, while $\delta$ incorporates a lower-bound constraint on the relevant operator quantities, and q controls the geometric relation between the vectors involved through the condition $⟨x,y⟩=q$. Thus, rather than being intended as a canonical choice among all possible extensions, the triple $(f,\delta ,q)$ is designed to combine these three established features in a single framework for operator sequences. The special cases listed in Section 2 demonstrate that this choice simultaneously recovers several previously studied numerical-radius-type notions.

The use of operator sequences, rather than only finite operator tuples, allows the proposed framework to capture the collective behavior of an entire sequence of operators. In this setting, the summability condition ${\sum }_{n=1}^{\infty }f(\parallel A_n\parallel )<\infty$ ensures that the contribution of the infinitely many coordinates remains controlled. Finite operator tuples can then be viewed as particular cases obtained by taking all but finitely many coordinates to be zero, whereas the sequential framework also accommodates genuinely infinite families. Thus, the proposed numerical radius is not merely a formal extension of finite-dimensional tuples, but provides a unified way to measure the overall numerical-radius behavior of an operator sequence.

This work is organized as follows: In Section 2, the basic definitions and preliminary results required throughout the paper are presented. In Section 3, several fundamental properties of the sequential joint $(f,\delta ,q)$-numerical radius function are established, and lower and upper bounds for this function are derived. Finally, Section 4 is devoted to the investigation of the sequential joint $(f,\delta ,q)$-numerical radius function for operator sequences whose coordinate operators are sectorial.

2. Basic Definitions and Preliminaries

In this section, the basic definitions and preliminary results required for the subsequent analysis are presented.

In order to extend the classical numerical radius concept to a more general setting, operator sequences and their associated operations are first introduced.

Definition 5. 

Let $\mathbb{A}={\left(A_n\right)}_{n=1}^{\infty }$ and $\mathbb{B}={\left(B_n\right)}_{n=1}^{\infty }$ be two operator sequences with $A_n,B_n\in B\left(H\right)$ for every $n\ge 1$, and let $\lambda \in \mathbb{C}$. The following notions are defined:

1.

$\mathbb{A}=0$ if and only if $A_n=0,$ for every $n\ge 1$.

2.

$\lambda \mathbb{A}={\left(\lambda A_n\right)}_{n=1}^{\infty }$.

3.

$\mathbb{A}+\mathbb{B}={(A_n+B_n)}_{n=1}^{\infty }$.

4.

$\mathbb{A}\mathbb{B}={\left(A_n{B}_n\right)}_{n=1}^{\infty }$.

5.

${\mathbb{A}}^{*}={\left(A_n^{*}\right)}_{n=1}^{\infty }$.

6.

$Re\mathbb{A}={\left(Re{A}_n\right)}_{n=1}^{\infty }$ and $Im\mathbb{A}={\left(Im{A}_n\right)}_{n=1}^{\infty }$.

7.

$\left|\mathbb{A}\right|={\left(|A_n|\right)}_{n=1}^{\infty }$.

8.

$\mathbb{A}x={\left(A_{n}x\right)}_{n=1}^{\infty }$ and $\parallel \mathbb{A}x\parallel ={\left({\displaystyle \sum _{n=1}^{\infty }}{\parallel A_{n}x\parallel }^2\right)}^{{\displaystyle \frac{1}{2}}},x\in H$.

9.

The operator norm of the operator sequence $\mathbb{A}={\left(A_n\right)}_{n=1}^{\infty }$ is defined as

$$\parallel \mathbb{A}\parallel =sup\left\{{\left({\displaystyle \sum _{n=1}^{\infty }}{\parallel A_{n}x\parallel }^2\right)}^{{\displaystyle \frac{1}{2}}}:x\in S^1\left(H\right)\right\}={∥{\displaystyle \sum _{n=1}^{\infty }}{A}_n^{*}{A}_n∥}^{1/2}.$$

10.

$\left|⟨\mathbb{A}x,x⟩\right|={\left({\displaystyle \sum _{n=1}^{\infty }}{\left|⟨A_{n}x,x⟩\right|}^2\right)}^{{\displaystyle \frac{1}{2}}},x\in H$.

Based on the above definition, the following function, referred to as a function-norm (although it is not necessarily a norm in the usual sense), is introduced. This allows the introduction of a new definition of the numerical radius, which will be considered throughout the paper.

Definition 6. 

Let $f:[0,\infty )\to [0,\infty )$ be a continuous strictly increasing surjective function. Consider a sequence of linear bounded operators $\mathbb{A}={\left(A_n\right)}_{n=1}^{\infty }$ on H satisfying the summability condition ${\displaystyle \sum _{n=1}^{\infty }}f\left(\parallel A_n\parallel \right)<\infty .$ Define the following:

1.

${\parallel \mathbb{A}x\parallel }_f=f^{-1}\left({\displaystyle \sum _{n=1}^{\infty }}f\left(\parallel A_{n}x\parallel \right)\right),x\in H$.

2.

${\parallel \mathbb{A}\parallel }_f=sup\left\{{\parallel \mathbb{A}x\parallel }_f:x\in S^1\left(H\right)\right\}$.

Using the above function-norm, the following generalized definition of the $\delta$-numerical radius is considered.

Definition 7. 

Let $dimH\ge 2,f:[0,\infty )\to [0,\infty )$ be a continuous strictly increasing surjective function. Let $\mathbb{A}={\left(A_n\right)}_{n=1}^{\infty }$ be a sequence of Hilbert space operators $A_n,n\ge 1$ in $B\left(H\right)$ such that ${\displaystyle \sum _{n=1}^{\infty }}f\left(\parallel A_n\parallel \right)<\infty .$ Let $q\in \mathbb{C}$ satisfy $\left|q\right|\le 1$ for each $0\le \delta \le {\parallel \mathbb{A}\parallel }_f.$ Define

1.

$$S_{(f,\delta )}^1(\mathbb{A},H)=\left\{(x,y)\in S^1\left(H\right)\times S^1{\left(H\right):\parallel \mathbb{A}x\parallel }_f\ge \delta ,{\parallel {\mathbb{A}}^{*}y\parallel }_f\ge \delta \right\}.$$

2.

$$S_{(f,\delta ,q)}^1(\mathbb{A},H)=\left\{(x,y)\in S^1\left(H\right)\times S^1{\left(H\right):\parallel \mathbb{A}x\parallel }_f\ge \delta ,{\parallel {\mathbb{A}}^{*}y\parallel }_f\ge \delta ,⟨x,y⟩=q\right\}.$$

3.

The set

$$W_{(f,\delta )}\left(\mathbb{A}\right)=\left\{⟨A_{n}x,y⟩\in {\mathbb{C}}^{\infty }:(x,y)\in S_{(f,\delta )}^1(\mathbb{A},H)\right\}$$

is called the sequential joint $(f,\delta )$-numerical range of $\mathbb{A}$.

4.

The set

$$W_{(f,\delta ,q)}\left(\mathbb{A}\right)=\left\{⟨A_{n}x,y⟩\in {\mathbb{C}}^{\infty }:(x,y)\in S_{(f,\delta ,q)}^1(\mathbb{A},H)\right\}$$

is called the sequential joint $(f,\delta ,q)$-numerical range of $\mathbb{A}$.

5.

For every $x,y\in H$, define

$${\left|⟨\mathbb{A}x,y⟩\right|}_f=f^{-1}\left({\displaystyle \sum _{n=1}^{\infty }}f\left(|⟨A_{n}x,y⟩|\right)\right).$$

6.

The non-negative number

$${\omega }_{(f,\delta )}\left(\mathbb{A}\right)=sup\left\{{\left|⟨\mathbb{A}x,y⟩\right|}_f:(x,y)\in S_{(f,\delta )}^1(\mathbb{A},H)\right\}$$

is called the sequential joint $(f,\delta )$-numerical radius of $\mathbb{A}$.

7.

The non-negative number

$${\omega }_{(f,\delta ,q)}\left(\mathbb{A}\right)=sup\left\{{\left|⟨\mathbb{A}x,y⟩\right|}_f:(x,y)\in S_{(f,\delta ,q)}^1(\mathbb{A},H)\right\}$$

is called the sequential joint $(f,\delta ,q)$-numerical radius of $\mathbb{A}$.

From a geometric viewpoint, when $\delta >0$, the condition set consists of pairs of unit vectors satisfying the constraint $⟨x,y⟩=q$ together with the lower-bound conditions imposed by $\delta$. Thus, the parameter $\delta$ restricts the admissible pairs of vectors by excluding those for which the relevant operator quantities are smaller than $\delta$. As $\delta$ increases, these restrictions become stronger, and the corresponding condition set becomes more constrained.

The introduced notion contains several existing numerical radius concepts as particular cases. The following observations demonstrate that the proposed definition provides a common framework for various known generalizations.

1.

If $\mathbb{A}=(A,0,\dots ),A\in B\left(H\right),\delta =0,q=1,$ then ${\omega }_{(f,0,1)}\left(\mathbb{A}\right)$ coincides with the classical numerical radius of the operator A (see [4,33]).

2.

If $\mathbb{A}=(A,0,\dots ),A\in B\left(H\right),0\le \delta \le \parallel A\parallel ,q=1,$ then ${\omega }_{(f,\delta ,1)}\left(\mathbb{A}\right)$ coincides with the $\delta$-numerical radius of the operator A (see [9]).

3.

If $\mathbb{A}=(A_1,A_2,0,\dots ),A_1,A_2=A_1^{*}{A}_1\in B\left(H\right),f\left(t\right)=t^2,0\le t<\infty ,\delta =0,q=1,$ then ${\omega }_{(t^2,0,1)}\left(\mathbb{A}\right)$ coincides with the Davis–Wielandt radius of the operator $A_1$ (see [34,35]).

4.

If $f\left(t\right)=t^2(t^p,1\le p<\infty ),0\le t<\infty ,\mathbb{A}=(A_1,\dots ,A_n,0,\dots ),n\in \mathbb{N},\delta =0,q=1,$ then ${\omega }_{(t^2,0,1)}\left(\mathbb{A}\right)\left({\omega }_{(t^p,0,1)}\left(\mathbb{A}\right)\right)$ coincides with the joint Euclidean numerical radius (joint p-numerical radius) of the n-tuple operators $\left(A_1,\dots ,A_n\right)\in B^n\left(H\right)$ (see [14]).

5.

If $f\left(t\right)=t^2,0\le t<\infty ,\mathbb{A}=(A_1,\dots ,A_n,0,\dots ),n\in \mathbb{N},0\le \delta \le \parallel \mathbb{A}\parallel =\underset{\parallel x\parallel =1}{sup}{\left({\displaystyle \sum _{m=1}^n}{\parallel A_{m}x\parallel }^2\right)}^{{\displaystyle \frac{1}{2}}},q=1,$ then ${\omega }_{(t^2,\delta ,1)}\left(\mathbb{A}\right)$ coincides with the $\delta$-numerical radius of the $n-$tuple operators $\left(A_1,\dots ,A_n\right)\in B^n\left(H\right)$ (see [36]).

6.

If $f\left(t\right)=t^2,0\le t<\infty ,\mathbb{A}=(A_1,\dots ,A_n,0,\dots ),n\in \mathbb{N},\delta =0,$ then ${\omega }_{(t^2,0,q)}\left(\mathbb{A}\right)$ coincides with the q-numerical radius of the n-tuple operator $\left(A_1,\dots ,A_n\right)\in B^n\left(H\right)$ (see [22]).

7.

In case when $\mathbb{A}=(A_1,\dots ,A_n,0,\dots ),n\in \mathbb{N},A_m\in B\left(H\right),1\le m\le n,\delta =0,q=1,$ the function numerical radius was previously considered in [37], where generalizations of known inequalities and structural results were obtained.

Remark 1. 

The results in [36,37] are closely related to the present work. While [36] deals with the joint δ-numerical radius of finite operator tuples and [37] considers an f-numerical radius for finite operator tuples, the present paper combines the parameters f, δ, and q within a unified framework for sequences of operators. Thus, the main novelty of the present work is the simultaneous integration of these features into the sequential joint $(f,\delta ,q)$-numerical radius and the resulting properties and inequalities.

Throughout this article, for $-\infty <\delta \le 0$, the $(f,\delta ,q)$-numerical radius is identified with its value at $\delta =0$, namely,

$${\omega }_{(f,\delta ,q)}\left(\mathbb{A}\right)={\omega }_{(f,0,q)}\left(\mathbb{A}\right),q\in \mathbb{C},\left|q\right|\le 1,$$

so that the consistency of the theory is preserved for nonpositive values of the parameter.

For any Hilbert space operators sequences $\mathbb{A}=\left(A_n\right),A_n\in B\left(H\right),n\ge 1$ and $\lambda \in \mathbb{R},q\in \mathbb{C}$ with $\left|q\right|\le 1,$ we define

$$S_{(f,\lambda ,q)}^1(\mathbb{A},H)=\left\{(x,y)\in S^1\left(H\right)\times S^1{\left(H\right):\parallel \mathbb{A}x\parallel }_f\ge \lambda ,{\parallel {\mathbb{A}}^{*}y\parallel }_f\ge \lambda ,⟨x,y⟩=q\right\}.$$

On the other hand, it is easy to verify that, for any sequence $\mathbb{A}={\left(A_n\right)}_{n=1}^{\infty }$ in $B\left(H\right)$ and any $q\in \mathbb{C}$ with $\left|q\right|\le 1$, the identity

$${\omega }_{(f,\delta ,q)}\left(\mathbb{A}\right)={\omega }_{(f,\delta ,\overline{q})}\left({\mathbb{A}}^{*}\right)$$

holds for every $0\le \delta \le {\parallel \mathbb{A}\parallel }_f$. Moreover, for every $\tau \le \delta \le {\parallel \mathbb{A}\parallel }_f$,

$${\omega }_{(f,\delta ,q)}\left(\mathbb{A}\right)\le {\omega }_{(f,\tau ,q)}\left(\mathbb{A}\right)\le {\omega }_{(f,0,q)}\left(\mathbb{A}\right)\le {\parallel \mathbb{A}\parallel }_f.$$

The following example illustrates the introduced concept in a concrete setting.

Example 1. 

Let $H={\mathbb{C}}^2$ and consider the following operator sequence

$$\mathbb{A}=\left(A_n\right),A_n=\left(\begin{array}{cc}{\alpha }^n& 0\\ 0& {\alpha }^n\end{array}\right):{\mathbb{C}}^2\to {\mathbb{C}}^2,n\ge 1,\alpha \in \mathbb{C},0<\left|\alpha \right|<1,$$

and assume that $f\left(t\right)=t^2,0\le t<\infty .$ It is clear that $A_{n}x={\alpha }^n(x_1,x_2),x=(x_1,x_2)\in {\mathbb{C}}^2$ and $\parallel A_n{x\parallel =\left|\alpha \right|}^n\parallel x\parallel$

$${\parallel \mathbb{A}x\parallel }_{t^2}={\left({\displaystyle \sum _{n=1}^{\infty }}{\left|\alpha \right|}^{2n}{\parallel x\parallel }^2\right)}^{{\displaystyle \frac{1}{2}}}={\displaystyle \frac{\left|\alpha \right|}{\sqrt{1-{\left|\alpha \right|}^2}}}\parallel x\parallel .$$

Hence,

$${\parallel \mathbb{A}\parallel }_{t^2}={\displaystyle \frac{\left|\alpha \right|}{\sqrt{1-{\left|\alpha \right|}^2}}}.$$

Then, for $0\le \delta \le {\displaystyle \frac{\left|\alpha \right|}{\sqrt{1-{\left|\alpha \right|}^2}}},$ from the relation

$${\parallel \mathbb{A}x\parallel }_{t^2}\ge \delta ,$$

it is obtained

$$\parallel x\parallel \ge {\displaystyle \frac{\delta \sqrt{1-{\left|\alpha \right|}^2}}{\left|\alpha \right|}},x=(x_1,x_2).$$

Similarly, since

$$A_n^{*}=\left(\begin{array}{cc}{\overline{\alpha }}^n& 0\\ 0& {\overline{\alpha }}^n\end{array}\right):{\mathbb{C}}^2\to {\mathbb{C}}^2,n\ge 1,\alpha \in \mathbb{C}$$

and ${\mathbb{A}}^{*}=\left(A_n^{*}\right)$,

$$\parallel y\parallel \ge {\displaystyle \frac{\delta \sqrt{1-{\left|\alpha \right|}^2}}{\left|\alpha \right|}},y=(y_1,y_2).$$

Hence, for $q\in \mathbb{C},\left|q\right|\le 1$

$$\begin{array}{cc}\hfill S_{(t^2,\delta ,q)}(\mathbb{A},{\mathbb{C}}^2)=& \left\{(x,y)\in S^1\left({\mathbb{C}}^2\right)\times S^1\left({\mathbb{C}}^2\right):\parallel x\parallel \ge {\displaystyle \frac{\delta \sqrt{1-{\left|\alpha \right|}^2}}{\left|\alpha \right|}},\parallel y\parallel \ge {\displaystyle \frac{\delta \sqrt{1-{\left|\alpha \right|}^2}}{\left|\alpha \right|}},⟨x,y⟩=q\right\}.\hfill \end{array}$$

Then, for any $0\le \delta \le {\displaystyle \frac{\left|\alpha \right|}{\sqrt{1-{\left|\alpha \right|}^2}}},$ it is obtained that

$$\begin{array}{cc}\hfill {\omega }_{(t^2,\delta ,q)}\left(\mathbb{A}\right)& =sup\left\{{\left|⟨\mathbb{A}x,y⟩\right|}_{t^2}:(x,y)\in S_{(t^2,\delta ,q)}^1\left({\mathbb{C}}^2\right)\right\}\hfill \\ \hfill & =sup\left\{{\left({\displaystyle \sum _{n=1}^{\infty }}{\left|\alpha \right|}^{2n}{\left|⟨x,y⟩\right|}^2\right)}^{{\displaystyle \frac{1}{2}}}:(x,y)\in S_{(t^2,\delta ,q)}^1\left({\mathbb{C}}^2\right)\right\}\hfill \\ \hfill & =sup\left\{{\displaystyle \frac{\left|\alpha \right|}{\sqrt{1-{\left|\alpha \right|}^2}}}\left|⟨x,y⟩\right|:(x,y)\in S_{(t^2,\delta ,q)}^1\left({\mathbb{C}}^2\right)\right\}\hfill \\ \hfill & ={\displaystyle \frac{\left|\alpha \right|}{\sqrt{1-{\left|\alpha \right|}^2}}}\left|q\right|,\left|q\right|\le 1.\hfill \end{array}$$

In the following, some additional assumptions on the function f will be considered, together with an auxiliary set that will be used in later discussions.

Throughout this article, $f:[0,\infty )\to [0,\infty )$ is assumed to be a continuous, strictly increasing, and surjective function. Under these assumptions, $f\left(0\right)=0$ is obtained. Moreover, whenever $f:[0,\infty )\to [0,\infty )$ is convex on $[0,\infty )$, the inequality

$$f\left(\alpha x\right)\le \alpha f\left(x\right)$$

is satisfied for every $0\le \alpha \le 1$ and $0\le x<\infty$. A function

$$f:[0,\infty )\to [0,\infty )$$

is said to be geometrically convex if the inequality

$$f\left(\sqrt{xy}\right)\le \sqrt{f\left(x\right)f\left(y\right)}$$

is satisfied for all $x,y\in [0,\infty )$.

Similarly, a function $f:[0,\infty )\to [0,\infty )$ is said to be multiplicative provided that

$$f\left(xy\right)=f\left(x\right)f\left(y\right)$$

holds for all $x,y\in [0,\infty )$ (see [38]).

In this case, $f\left(0\right)=0$ or $f\left(0\right)=1$ and $f\left(1\right)=0$ or $f\left(1\right)=1.$

The choice of the function f affects the tightness of the derived bounds, since different choices of f lead to different scalarizations and consequently to different estimates.

3. Sequential Joint $(\mathit{f},\mathit{\delta },\mathit{q})$-Numerical Radius

In this section, several fundamental properties of the sequential joint $(f,\delta ,q)$-numerical radius function associated with sequences of bounded linear operators on a Hilbert space are first examined. Subsequently, generalized formulas for the lower and upper bounds of this numerical radius function are established.

Theorem 1. 

Let f be a convex function and $\mathbb{A}={\left(A_n\right)}_{n=1}^{\infty }$ be a sequence of Hilbert space operators, where $A_n\in B\left(H\right)$ for every $n\ge 1$, and let $q\in \mathbb{C}$ satisfy $\left|q\right|\le 1$. Then, for $0\le \delta \le {inf}_{n\ge 1}\parallel A_n\parallel ,$

$$\underset{n\ge 1}{sup}{\omega }_{(f,\delta ,q)}\left(A_n\right)\le {\omega }_{(f,\delta ,q)}\left(\mathbb{A}\right)\le {\displaystyle \sum _{n=1}^{\infty }}{\omega }_q\left(A_n\right).$$

Proof. 

Let $x,y\in H$ be arbitrary. By monotonicity of the function $f^{-1},$ we have

$$|⟨A_{n}x,y⟩|=f^{-1}f\left(|⟨A_{n}x,y⟩|\right)\le f^{-1}\left({\displaystyle \sum _{n=1}^{\infty }}f\left(|⟨A_{n}x,y⟩|\right)\right).$$

(4)

Hence, for every $x\in H,$

$$\parallel A_{n}x\parallel =f^{-1}\left(f\left(\parallel A_{n}x\parallel \right)\right)\le f^{-1}\left({\displaystyle \sum _{n=1}^{\infty }}f\left(\parallel A_{n}x\parallel \right)\right)$$

which immediately yields

$$\parallel A_n{x\parallel \le \parallel \mathbb{A}x\parallel }_f,n\ge 1.$$

In the same manner, for every $y\in H,$

$$\parallel A_n^{*}y\parallel =f^{-1}\left(f\left(\parallel A_n^{*}y\parallel \right)\right)\le f^{-1}\left({\displaystyle \sum _{n=1}^{\infty }}f\left(\parallel A_n^{*}y\parallel \right)\right),n\ge 1$$

and therefore

$$\parallel A_n^{*}y\parallel \le \parallel {\mathbb{A}}^{*}{y\parallel }_f,n\ge 1.$$

Consequently,

$$S_{(f,\delta ,q)}^1(A_n,H)\subset S_{(f,\delta ,q)}^1(\mathbb{A},H),n\ge 1.$$

(5)

Combining (4) and (5) it is obtained that

$${\omega }_{(f,\delta ,q)}\left(A_n\right)\le {\omega }_{(f,\delta ,q)}\left(\mathbb{A}\right),n\ge 1.$$

Taking the supremum over all n gives

$$\underset{n\ge 1}{sup}{\omega }_{(f,\delta ,q)}\left(A_n\right)\le {\omega }_{(f,\delta ,q)}\left(\mathbb{A}\right).$$

On the other hand, recall that the convexity of f implies that its inverse function $f^{-1}$ is concave. Hence, by [37]

$$f^{-1}(a+b)\le f^{-1}\left(a\right)+f^{-1}\left(b\right),a,b\ge 0.$$

Moreover, the superadditivity of f [37] ensures that

$${\displaystyle \sum _{m=1}^n}f\left(a_m\right)\le f\left({\displaystyle \sum _{m=1}^n}{a}_m\right),$$

for every $n\ge 1$ and $a_m\ge 0,m=1,\dots ,n.$

Since $f^{-1}$ is monotonically increasing, it follows that

$$f^{-1}\left({\displaystyle \sum _{m=1}^n}f\left(a_m\right)\right)\le {\displaystyle \sum _{m=1}^n}{a}_m,$$

for every $n\ge 1$ and $a_m\ge 0,m=1,\dots ,n.$

Passing to the limit $n\to \infty$ and using the continuity of $f^{-1}$, it follows that

$$f^{-1}\left({\displaystyle \sum _{m=1}^{\infty }}f\left(a_m\right)\right)\le {\displaystyle \sum _{m=1}^{\infty }}{a}_m,$$

for $a_m\ge 0,m\ge 1.$

Applying this inequality with $a_m=\left|⟨A_{m}x,y⟩\right|,m\ge 1$ we deduce that

$$f^{-1}\left({\displaystyle \sum _{n=1}^{\infty }}f\left(|⟨A_{n}x,y⟩|\right)\right)\le {\displaystyle \sum _{n=1}^{\infty }}\left|⟨A_{n}x,y⟩\right|,$$

(6)

for every $x,y\in H.$

Furthermore, it is easily to see that

$$S_{(f,\delta ,q)}^1(\mathbb{A},H)\subset S_q^1\left(H\right).$$

(7)

Therefore, from (6) and (7) it is established that

$${\omega }_{(f,\delta ,q)}\left(\mathbb{A}\right)\le {\displaystyle \sum _{n=1}^{\infty }}{\omega }_q\left(A_n\right).$$

This completes the proof. □

Theorem 2. 

Let f be a multiplicative function and let $\mathbb{A}=(A_1,\dots ,A_n,0,\dots ),\mathbb{B}=(B_1,\dots ,B_n,0,\dots ),$ where $A_m,B_m\in B\left(H\right),$ for each $m=1,\dots ,n.$ Then, for $\delta =0$ and every $q\in \left\{z\in \mathbb{C}:0<\left|z\right|\le 1\right\}$

(1) 

${\parallel \mathbb{A}\mathbb{B}\parallel }_f\le {\parallel \mathbb{A}\parallel }_f{\parallel \mathbb{B}\parallel }_f,$

(2) 

${\displaystyle \frac{\left|q\right|}{2f^{-1}\left(n\right)}}{\parallel \mathbb{A}\parallel }_f\le {\omega }_{(f,0,q)}\left(\mathbb{A}\right)\le {\parallel \mathbb{A}\parallel }_f,$

(3) 

${\omega }_{(f,0,q)}\left(\mathbb{A}\mathbb{B}\right)\le {\displaystyle \frac{4{\left(f^{-1}\left(n\right)\right)}^2}{{\left|q\right|}^2}}{\omega }_{(f,0,q)}\left(\mathbb{A}\right){\omega }_{(f,0,q)}\left(\mathbb{B}\right)$ (polarization formula).

Proof. (1)

Let $x\in S^1\left(H\right).$ Then

$$\begin{array}{cc}\hfill f^{-1}\left({\displaystyle \sum _{m=1}^n}f\left(\parallel A_m{B}_{m}x\parallel \right)\right)& \le f^{-1}\left({\displaystyle \sum _{m=1}^n}f\left(\parallel A_m\parallel \parallel B_{m}x\parallel \right)\right)\hfill \\ \hfill & \le f^{-1}\left({\displaystyle \sum _{m=1}^n}f\left(\parallel A_m\parallel \right)f\left(\parallel B_{m}x\parallel \right)\right)\hfill \\ \hfill & \le f^{-1}\left(\underset{1\le m\le n}{sup}f\left(\parallel A_m\parallel \right)\right)f^{-1}\left({\displaystyle \sum _{m=1}^n}f\left(\parallel B_{m}x\parallel \right)\right)\hfill \\ \hfill & =f^{-1}\left(f\left(\underset{z\in S^1\left(H\right)}{sup}\underset{1\le m\le n}{sup}\parallel A_{m}z\parallel \right)\right)f^{-1}\left({\displaystyle \sum _{m=1}^n}f\left(\parallel B_{m}x\parallel \right)\right)\hfill \\ \hfill & \le \underset{z\in S^1\left(H\right)}{sup}{f}^{-1}\left({\displaystyle \sum _{m=1}^n}f\left(\parallel A_{m}z\parallel \right)\right)f^{-1}\left({\displaystyle \sum _{m=1}^n}f\left(\parallel B_{m}x\parallel \right)\right).\hfill \end{array}$$

Taking the supremum over all $x\in S^1\left(H\right),$ we conclude that

$${\parallel \mathbb{A}\mathbb{B}\parallel }_f\le {\parallel \mathbb{A}\parallel }_f{\parallel \mathbb{B}\parallel }_f.$$

(2)

Let $x,y\in S^1\left(H\right).$ Then

$$\left|⟨\mathbb{A}x,y⟩\right|=f^{-1}\left({\displaystyle \sum _{m=1}^n}f\left(|⟨A_{m}x,y⟩|\right)\right)\le f^{-1}\left({\displaystyle \sum _{m=1}^n}f\left(\parallel A_{m}x\parallel \right)\right).$$

Hence

$${\omega }_{(f,0,q)}\left(\mathbb{A}\right)\le {\parallel \mathbb{A}\parallel }_f.$$

On the other hand, since f is multiplicative, we use the property

$$f^{-1}\left(ab\right)=f^{-1}\left(a\right)f^{-1}\left(b\right),a,b\ge 0.$$

Then, for every $x\in S^1\left(H\right)$

$$\begin{array}{cc}\hfill f^{-1}\left({\displaystyle \sum _{m=1}^n}f\left(\parallel A_{m}x\parallel \right)\right)& \le f^{-1}\left(nf\left(\underset{1\le m\le n}{sup}\parallel A_{m}x\parallel \right)\right)\hfill \\ \hfill & \le f^{-1}\left(n\right)f^{-1}\left(f\left(\underset{1\le m\le n}{sup}\parallel A_{m}x\parallel \right)\right)\hfill \\ \hfill & =f^{-1}\left(n\right)\underset{1\le m\le n}{sup}\parallel A_{m}x\parallel \hfill \\ \hfill & \le f^{-1}\left(n\right)\underset{1\le m\le n}{sup}\parallel A_m\parallel .\hfill \end{array}$$

Therefore,

$${\parallel \mathbb{A}\parallel }_f\le f^{-1}\left(n\right)\underset{1\le m\le n}{sup}\parallel A_m\parallel .$$

Combining this estimate with the inequality established in [31], it is obtained that

$$\begin{array}{cc}\hfill {\parallel \mathbb{A}\parallel }_f& \le {\displaystyle \frac{2f^{-1}\left(n\right)}{\left|q\right|}}\underset{1\le m\le n}{sup}{\omega }_q\left(A_m\right)\hfill \\ \hfill & \le {\displaystyle \frac{2f^{-1}\left(n\right)}{\left|q\right|}}\underset{1\le m\le n}{sup}\underset{(x,y)\in S_q^1\left(H\right)}{sup}\left|⟨A_{m}x,y⟩\right|\hfill \\ \hfill & \le {\displaystyle \frac{2f^{-1}\left(n\right)}{\left|q\right|}}\underset{(x,y)\in S_q^1\left(H\right)}{sup}\underset{1\le m\le n}{sup}\left|⟨A_{m}x,y⟩\right|\hfill \\ \hfill & \le {\displaystyle \frac{2f^{-1}\left(n\right)}{\left|q\right|}}\underset{(x,y)\in S_q^1\left(H\right)}{sup}{f}^{-1}\left({\displaystyle \sum _{m=1}^n}f\left(|⟨A_{m}x,y⟩|\right)\right)\hfill \\ \hfill & ={\displaystyle \frac{2f^{-1}\left(n\right)}{\left|q\right|}}{\omega }_{(f,0,q)}\left(\mathbb{A}\right).\hfill \end{array}$$

Consequently,

$${\displaystyle \frac{\left|q\right|}{2f^{-1}\left(n\right)}}{\parallel \mathbb{A}\parallel }_f\le {\omega }_{(f,0,q)}\left(\mathbb{A}\right)\le {\parallel \mathbb{A}\parallel }_f.$$

(3)

Combining assertions (1) and (2) of the theorem, we obtain

$${\omega }_{(f,0,q)}\left(\mathbb{A}\mathbb{B}\right)\le {\parallel \mathbb{A}\mathbb{B}\parallel }_f\le {\parallel \mathbb{A}\parallel }_f{\parallel \mathbb{B}\parallel }_f\le {\displaystyle \frac{4{\left(f^{-1}\left(n\right)\right)}^2}{{\left|q\right|}^2}}{\omega }_{(f,0,q)}\left(\mathbb{A}\right){\omega }_{(f,0,q)}\left(\mathbb{B}\right),$$

which completes the proof.

□

Remark 2. 

If the function f is not multiplicative, then assertions (2) and (3) of Theorem 2 become

$${\displaystyle \frac{\left|q\right|}{2n}}{\parallel \mathbb{A}\parallel }_f\le {\omega }_{(f,0,q)}\left(\mathbb{A}\right),$$

and

$${\omega }_{(f,0,q)}\left(\mathbb{A}\mathbb{B}\right)\le {\displaystyle \frac{4n^2}{{\left|q\right|}^2}}{\omega }_{(f,0,q)}\left(\mathbb{A}\right){\omega }_{(f,0,q)}\left(\mathbb{B}\right),$$

respectively.

Corollary 1. 

Let

$$\mathbb{A}=(A_1,\dots ,A_n,0,\dots ),A_m\in B\left(H\right),1\le m\le n.$$

If $\delta =0$ and $q=1$, then assertions (1) and (3) of Theorem 2 reduce to the corresponding results established in [39], while assertion (2) reduces to a well-known joint numerical radius inequality; see [40].

As a consequence of Theorem 2, we obtain the following corollary. Assertion (1) recovers the result in [22], whereas assertion (2) recovers the result in [40].

Corollary 2. 

Let $f\left(t\right)=t^2$ for $t\ge 0.$

(1) 

If $0<q<1,$ then

$${\displaystyle \frac{q}{2\sqrt{n}}}{\parallel \mathbb{A}\parallel }_{t^2}\le {\omega }_{(t^2,0,q)}\left(\mathbb{A}\right)\le {\parallel \mathbb{A}\parallel }_{t^2}.$$

(2) 

If $q=1,$ then

$${\displaystyle \frac{1}{2\sqrt{n}}}{\parallel \mathbb{A}\parallel }_{t^2}\le {\omega }_{(t^2,0,1)}\left(\mathbb{A}\right)\le {\parallel \mathbb{A}\parallel }_{t^2}.$$

Theorem 3. 

Let $\mathbb{A}={\left(A_n\right)}_{n=1}^{\infty }$ and $\mathbb{B}={\left(B_n\right)}_{n=1}^{\infty }$ be two operators sequences on $B\left(H\right).$ Then the following assertions hold.

(1) 

If f is multiplicative, then for every $\lambda \in \mathbb{C},\lambda \ne 0$ and $0\le \delta \le {\parallel \lambda \mathbb{A}\parallel }_f,$

$${\omega }_{(f,\delta ,q)}\left(\lambda \mathbb{A}\right)=\left|\lambda \right|{\omega }_{\left(f,{\displaystyle \frac{\delta }{\left|\lambda \right|}},q\right)}\left(\mathbb{A}\right).$$

(2) 

Assume that f is a geometrically convex function. Then, for any operator sequences $\mathbb{A}={\left(A_n\right)}_{n=1}^{\infty }$ and $\mathbb{B}={\left(B_n\right)}_{n=1}^{\infty }$ on $B\left(H\right),$ and for every $0\le \delta \le {\parallel \mathbb{A}+\mathbb{B}\parallel }_f$ the following inequality holds:

$${\omega }_{(f,\delta ,q)}(\mathbb{A}+\mathbb{B})\le {\omega }_{(f,\delta -\parallel \mathbb{B}{\parallel }_f,q)}\left(\mathbb{A}\right)+{\omega }_{(f,\delta -\parallel \mathbb{A}{\parallel }_f,q)}\left(\mathbb{B}\right).$$

In particular, if $\alpha \le max\left\{{\parallel \mathbb{A}\parallel }_f,{\parallel \mathbb{B}\parallel }_f\right\},$ then

$${\omega }_{(f,\delta ,q)}(\mathbb{A}+\mathbb{B})\le {\omega }_{(f,\delta -\alpha ,q)}\left(\mathbb{A}\right)+{\omega }_{(f,\delta -\alpha ,q)}\left(\mathbb{B}\right).$$

(3) 

For every operator sequence $\mathbb{A}={\left(A_n\right)}_{n=1}^{\infty }$ on $B\left(H\right)$ and every $0\le \delta \le {\parallel \mathbb{A}\parallel }_f,$ the following equality holds:

$${\omega }_{(f,\delta ,q)}\left(\mathbb{A}\right)={\omega }_{(f,\delta ,\overline{q})}\left({\mathbb{A}}^{*}\right).$$

Moreover, if $\mathbb{A}={\mathbb{A}}^{*},$ then for any $0\le \delta \le {\parallel \mathbb{A}\parallel }_f$

$${\omega }_{(f,\delta ,q)}\left(\mathbb{A}\right)={\omega }_{(f,\delta ,\overline{q})}\left(\mathbb{A}\right).$$

Finally, if $q=\overline{q},$ then for any $0\le \delta \le {\parallel \mathbb{A}\parallel }_f$ is true

$${\omega }_{(f,\delta ,q)}\left(\mathbb{A}\right)={\omega }_{(f,\delta ,q)}\left({\mathbb{A}}^{*}\right).$$

Proof. (1)

For any $x,y\in H,$ we have

$$\begin{array}{cc}\hfill {\left|⟨\lambda \mathbb{A}x,y⟩\right|}_f=& f^{-1}\left({\displaystyle \sum _{m=1}^{\infty }}f\left(|⟨\lambda A_{n}x,y⟩|\right)\right)\hfill \\ \hfill & =f^{-1}\left({\displaystyle \sum _{n=1}^{\infty }}f\left(\left|\lambda \right||⟨A_{n}x,y⟩|\right)\right)\hfill \\ \hfill & =f^{-1}\left(f\left(\right|\lambda \left|\right){\displaystyle \sum _{n=1}^{\infty }}f\left(|⟨A_{n}x,y⟩|\right)\right)\hfill \\ \hfill & =\left|\lambda \right|f^{-1}\left({\displaystyle \sum _{n=1}^{\infty }}f\left(|⟨A_{n}x,y⟩|\right)\right).\hfill \end{array}$$

(8)

Moreover, it follows directly from the definition of the corresponding set that

$$S_{(f,\delta ,q)}^1(\lambda \mathbb{A},H)=S_{\left(f,{\displaystyle \frac{\delta }{\left|\lambda \right|}},q\right)}(\mathbb{A},H),\lambda \in \mathbb{C}\setminus \left\{0\right\}.$$

(9)

Therefore, combining (8) and (9), we obtain

$${\omega }_{(f,\delta ,q)}\left(\lambda \mathbb{A}\right)=\left|\lambda \right|{\omega }_{\left(f,{\displaystyle \frac{\delta }{\left|\lambda \right|}},q\right)}\left(\mathbb{A}\right).$$

(2)

Since f is monotonically increasing, then for every $x,y\in H,$

$${\displaystyle \sum _{n=1}^{\infty }}f\left(|⟨(A_n+B_n)x,y⟩|\right)\le {\displaystyle \sum _{n=1}^{\infty }}f\left(|⟨A_{n}x,y⟩|+|⟨B_{n}x,y⟩|\right).$$

Using the geometrically convexity of f (see [41]), we obtain

$$f^{-1}\left({\displaystyle \sum _{n=1}^{\infty }}f\left(|⟨(A_n+B_n)x,y⟩|\right)\right)\le f^{-1}\left({\displaystyle \sum _{n=1}^{\infty }}f\left(|⟨A_{n}x,y⟩|\right)\right)+f^{-1}\left({\displaystyle \sum _{n=1}^{\infty }}f\left(|⟨B_{n}x,y⟩|\right)\right).$$

Hence, for $x,y\in H,$

$${\left|⟨(\mathbb{A}+\mathbb{B})x,y⟩\right|}_f\le {\left|⟨\mathbb{A}x,y⟩\right|}_f+{\left|⟨\mathbb{B}x,y⟩\right|}_f.$$

(10)

Then,

$${\parallel (\mathbb{A}+\mathbb{B})x\parallel }_f\le {\parallel \mathbb{A}x\parallel }_f+{\parallel \mathbb{B}\parallel }_f{\mathrm{and}\parallel (\mathbb{A}+\mathbb{B})x\parallel }_f\le {\parallel \mathbb{A}\parallel }_f+{\parallel \mathbb{B}x\parallel }_f,x\in H.$$

Similarly, by the same technique it can be shown

$${\parallel (\mathbb{A}+\mathbb{B})}^{*}{y\parallel }_f\le \parallel {\mathbb{A}}^{*}{y\parallel }_f+\parallel {\mathbb{B}}^{*}{\parallel }_f{\mathrm{and}\parallel (\mathbb{A}+\mathbb{B})}^{*}{y\parallel }_f\le \parallel {\mathbb{A}}^{*}{\parallel }_f+{\parallel {\mathbb{B}}^{*}y\parallel }_f,y\in H.$$

Therefore, if $(x,y)\in S_{(f,\delta ,q)}^1(\mathbb{A}+\mathbb{B},H),$ then

$$\delta \le {\parallel (\mathbb{A}+\mathbb{B})x\parallel }_f\le {\parallel \mathbb{A}x\parallel }_f+{\parallel \mathbb{B}\parallel }_f\mathrm{and}\delta \le {\parallel (\mathbb{A}+\mathbb{B})x\parallel }_f\le {\parallel \mathbb{A}\parallel }_f+{\parallel \mathbb{B}x\parallel }_f,x\in H,$$

and

$${\delta \le \parallel (\mathbb{A}+\mathbb{B})}^{*}{y\parallel }_f\le \parallel {\mathbb{A}}^{*}{y\parallel }_f+\parallel {\mathbb{B}}^{*}{\parallel }_f{\mathrm{and}\delta \le \parallel (\mathbb{A}+\mathbb{B})}^{*}{y\parallel }_f\le \parallel {\mathbb{A}}^{*}{\parallel }_f+{\parallel {\mathbb{B}}^{*}y\parallel }_f,y\in H.$$

Thus, it is obtained that

$$S_{(f,\delta ,q)}^1(\mathbb{A}+\mathbb{B},H)\subset S_{(f,\delta -\parallel \mathbb{B}\parallel ,q)}^1(\mathbb{A},H)\cap S_{(f,\delta -\parallel \mathbb{A}\parallel ,q)}^1(\mathbb{B},H).$$

(11)

Consequently, from (10) and (11) it follows that

$${\omega }_{(f,\delta ,q)}(\mathbb{A}+\mathbb{B})\le {\omega }_{(f,\delta -\parallel \mathbb{B}{\parallel }_f,q)}\left(\mathbb{A}\right)+{\omega }_{(f,\delta -\parallel \mathbb{A}{\parallel }_f,q)}\left(\mathbb{B}\right),$$

for every $0\le \delta \le {\parallel \mathbb{A}+\mathbb{B}\parallel }_f.$

(3)

Let $(x,y)\in S_{(f,\delta ,q)}^1(\mathbb{A},H).$ Then

$${\left|⟨\mathbb{A}x,y⟩\right|}_f=f^{-1}\left({\displaystyle \sum _{n=1}^{\infty }}f\left(|⟨A_{n}x,y⟩|\right)\right)=f^{-1}\left({\displaystyle \sum _{n=1}^{\infty }}f\left(|⟨A_n^{*}y,x⟩|\right)\right)={\left|⟨{\mathbb{A}}^{*}y,x⟩\right|}_f.$$

Furthermore, by the definition of the set $S_{(f,\delta ,q)}^1(\mathbb{A},H),$

$$\begin{array}{cc}\hfill S_{(f,\delta ,q)}^1(\mathbb{A},H)& =\left\{(x,y)\in S^1\left(H\right)\times S^1\left(H\right):\parallel \mathbb{A}x\parallel \ge \delta ,\parallel {\mathbb{A}}^{*}y\parallel \ge \delta ,⟨x,y⟩=q\right\}\hfill \\ \hfill & =\left\{(y,x)\in S^1\left(H\right)\times S^1\left(H\right):\parallel {\mathbb{A}}^{*}y\parallel \ge \delta ,\parallel {\mathbb{A}}^{**}x\parallel =\parallel \mathbb{A}x\parallel \ge \delta ,⟨y,x⟩=\overline{q}\right\}\hfill \\ \hfill & =S_{(f,\delta ,\overline{q})}^1({\mathbb{A}}^{*},H).\hfill \end{array}$$

(12)

Hence, from (12) and the definition of $(f,\delta ,q)$-numerical radius

$${\omega }_{(f,\delta ,q)}\left(\mathbb{A}\right)={\omega }_{(f,\delta ,\overline{q})}\left({\mathbb{A}}^{*}\right).$$

This completes the proof. □

Theorem 3 contains several known results as special cases. In particular, we have the following consequences.

Corollary 3. (1)

If $\mathbb{A}=(A,0,\dots ),A\in B\left(H\right)$ and $\delta =0,q=1,$ then the result reduces to the result obtained in [33,42].

(2) 

If $f\left(t\right)=t^2,$ for $t\ge 0,\mathbb{A}=(A_1,\dots ,A_n,0,\dots ),A_m\in B\left(H\right),1\le m\le n$ and $\delta =0,q=1,$ then Theorem 3 recovers the result established in [39,43].

(3) 

If $f\left(t\right)=t^p,$ for $t\ge 0,$ and $1\le p,\mathbb{A}=(A_1,\dots ,A_n,0,\dots ),A_m\in B\left(H\right),1\le m\le n$ and $\delta =0,q=1,$ then the theorem yields the result obtained in [14].

(4) 

If $f:[0,\infty )\to [0,\infty )$ is a continuous increasing function and $\mathbb{A}=(A_1,\dots ,A_n,0,\dots )$ on $B\left(H\right)$ and $\delta =0,q=1,$ then the theorem has been established in [37].

(5) 

If $f\left(t\right)=t^2,$ for $t\ge 0,\mathbb{A}=(A_1,\dots ,A_n,0,\dots )$ on $B\left(H\right),$ and $0\le \delta \le \parallel \mathbb{A}\parallel ,q=1,$ then the theorem recovers the result obtained in [36].

(6) 

If $f\left(t\right)=t,$ for $t\ge 0,\mathbb{A}=(A,A^{*}A,0,\dots )$ on $B\left(H\right),$ and $\delta =0,q=1,$ then the theorem reduces to the corresponding result for the Davis–Wielandt radius obtained in [34,35].

(7) 

If $\mathbb{A}=(A,0,\dots )$ on $B\left(H\right)$ and $\delta =0,q\in \mathbb{C},\left|q\right|\le 1,$ then the theorem reduces to the corresponding result in [44].

Now, we provide an useful auxiliary lemma.

Lemma 1 

([45]). If $A\in B\left(H\right),A\ge 0,$ then for any $x\in H$,

$${\parallel Ax\parallel }^2\le \parallel A\parallel ⟨Ax,x⟩.$$

Theorem 4. 

Let f be a geometrically convex function. Then, for every operator sequence $\mathbb{A}={\left(A_n\right)}_{n=1}^{\infty }$ with $A_n\in B\left(H\right),n\ge 1,$ and $0\le \delta \le {\parallel \mathbb{A}\parallel }_f,$ the following inequality hold

$$f^2\left({\omega }_{(f,\delta ,q)}\left(\mathbb{A}\right)\right)\le f\left({\omega }_{(f,\delta )}\left(\right|\mathbb{A}\left|\right)\right)f\left({\omega }_{(f,\delta )}\left(\right|{\mathbb{A}}^{*}\left|\right)\right),$$

where $q\in \mathbb{C},\left|q\right|\le 1.$

Proof. 

By Lemma 1, for any $x,y\in H,$ we have

$$\begin{array}{cc}\hfill {\displaystyle \sum _{n=1}^{\infty }}f\left(|⟨A_{n}x,y⟩|\right)& \le {\displaystyle \sum _{n=1}^{\infty }}f\left(⟨|A_n{|x,x⟩}^{{\displaystyle \frac{1}{2}}}⟨|A_n^{*}{|y,y⟩}^{{\displaystyle \frac{1}{2}}}\right)\hfill \\ \hfill & \le {\displaystyle \sum _{n=1}^{\infty }}{f}^{{\displaystyle \frac{1}{2}}}\left(⟨\left|A_n\right|x,x⟩\right)f^{{\displaystyle \frac{1}{2}}}\left(⟨\left|A_n^{*}\right|y,y⟩\right)\hfill \\ \hfill & \le {\left({\displaystyle \sum _{n=1}^{\infty }}f\left(⟨\left|A_n\right|x,x⟩\right)\right)}^{{\displaystyle \frac{1}{2}}}{\left({\displaystyle \sum _{n=1}^{\infty }}f\left(⟨\left|A_n^{*}\right|y,y⟩\right)\right)}^{{\displaystyle \frac{1}{2}}}.\hfill \end{array}$$

(13)

On the other hand, for any $x,y\in H$ and $0\le \delta \le {\parallel \mathbb{A}\parallel }_f,$ we obtain

$$\begin{array}{cc}\hfill \delta \le {\parallel \mathbb{A}x\parallel }_f& =f^{-1}\left({\displaystyle \sum _{n=1}^{\infty }}f\left(\parallel A_{n}x\parallel \right)\right)\hfill \\ \hfill & =f^{-1}\left({\displaystyle \sum _{n=1}^{\infty }}f\left({⟨A_n^{*}{A}_{n}x,x⟩}^{{\displaystyle \frac{1}{2}}}\right)\right)\hfill \\ \hfill & =f^{-1}\left({\displaystyle \sum _{n=1}^{\infty }}f\left(\parallel |A_n|x\parallel \right)\right)\hfill \\ \hfill & ={\parallel \left|\mathbb{A}\right|x\parallel }_f.\hfill \end{array}$$

Hence, it is obtained that

$$S_{(f,\delta )}^1(\mathbb{A},H)\subset S_{(f,\delta )}^1\left(\right|\mathbb{A}|,H).$$

(14)

Similarly, it can be shown that for each $y\in S^1\left(H\right)$

$$\begin{array}{c}\hfill \delta \le \parallel {\mathbb{A}}^{*}{y\parallel }_f=f^{-1}\left({\displaystyle \sum _{n=1}^{\infty }}f\left(\parallel |A_n^{*}|y\parallel \right)\right)=\parallel |{\mathbb{A}}^{*}{|y\parallel }_f.\end{array}$$

Thus,

$$S_{(f,\delta )}^1({\mathbb{A}}^{*},H)\subset S_{(f,\delta )}^1\left(\right|{\mathbb{A}}^{*}|,H)$$

(15)

Therefore, from (13)–(15), it is established that for every $q\in \mathbb{C},$ with $\left|q\right|\le 1$

$$\begin{array}{cc}\hfill f^2\left({\left|⟨\mathbb{A}x,y⟩\right|}_f\right)& \le f\left({⟨\left|\mathbb{A}\right|x,x⟩}_f\right)f\left({⟨\left|\mathbb{A}\right|}^{*}{y,y⟩}_f\right)\hfill \\ \hfill & \le \underset{x\in S_{(f,\delta )}^1(\mathbb{A},H)}{sup}f\left({⟨\left|\mathbb{A}\right|x,x⟩}_f\right)\underset{y\in S_{(f,\delta )}^1({\mathbb{A}}^{*},H)}{sup}f\left(⟨|{\mathbb{A}}^{*}{|y,y⟩}_f\right)\hfill \\ \hfill & \le \underset{x\in S_{(f,\delta )}^1\left(\right|\mathbb{A}|,H)}{sup}f\left({⟨\left|\mathbb{A}\right|x,x⟩}_f\right)\underset{y\in S_{(f,\delta )}^1\left(\right|{\mathbb{A}}^{*}|,H)}{sup}f\left(⟨|{\mathbb{A}}^{*}{|y,y⟩}_f\right)\hfill \\ \hfill & =f\left({\omega }_{(f,\delta )}\left(\right|\mathbb{A}\left|\right)\right)f\left({\omega }_{(f,\delta )}\left(\right|{\mathbb{A}}^{*}\left|\right)\right).\hfill \end{array}$$

Then, for every $0\le \delta \le {\parallel \mathbb{A}\parallel }_f$ and $q\in \mathbb{C}$ with $\left|q\right|\le 1$

$$\underset{(x,y)\in S_{(f,\delta ,q)}^1(\mathbb{A},H)}{sup}{f}^2\left({⟨\left|\mathbb{A}\right|x,x⟩}_f\right)\le f\left({\omega }_{(f,\delta )}\left(\right|\mathbb{A}\left|\right)\right)f\left({\omega }_{(f,\delta )}\left(\right|{\mathbb{A}}^{*}\left|\right)\right).$$

This completes the proof. □

Theorem 5. 

Let f be a geometrically convex function and let $\mathbb{A}={\left(A_n\right)}_{n=1}^{\infty }$ be an operator sequence on $B\left(H\right).$ Then for any $0\le \delta \le {\parallel \mathbb{A}\parallel }_f,$

$${\omega }_{(f,\delta ,q)}\left(\mathbb{A}\right)\le f\left(1\right){\omega }_{\left(\sqrt{f},f^{-1}\left({\displaystyle \frac{f^2\left(\delta \right)}{f\left(1\right)}}\right)\right)}\left({\mathbb{A}}^{*}\mathbb{A}\right),$$

where $q\in \mathbb{C}$ satisfies $\left|q\right|\le 1.$

Proof. 

Since f is geometrically convex, for any $x,y\in S^1\left(H\right)$

$$\begin{array}{cc}\hfill f\left({\left|⟨\mathbb{A}x,y⟩\right|}_f\right)& ={\displaystyle \sum _{n=1}^{\infty }}f\left(|⟨A_{n}x,y⟩|\right)\hfill \\ \hfill & \le {\displaystyle \sum _{n=1}^{\infty }}f\left(\parallel A_{n}x\parallel \right)\hfill \\ \hfill & ={\displaystyle \sum _{n=1}^{\infty }}f\left({⟨A_n^{*}{A}_{n}x,x⟩}^{{\displaystyle \frac{1}{2}}}\right)\hfill \\ \hfill & ={\displaystyle \sum _{n=1}^{\infty }}f\left(\sqrt{1}\sqrt{⟨A_n^{*}{A}_{n}x,x⟩}\right)\hfill \\ \hfill & \le {\displaystyle \sum _{n=1}^{\infty }}\sqrt{f\left(1\right)}\sqrt{f\left(⟨A_n^{*}{A}_{n}x,x⟩\right)}\hfill \\ \hfill & \le \sqrt{f\left(1\right)}{f}^{{\displaystyle \frac{1}{2}}}\left(f^{-{\displaystyle \frac{1}{2}}}\left({\displaystyle \sum _{n=1}^{\infty }}{f}^{{\displaystyle \frac{1}{2}}}\left(⟨A_n^{*}{A}_{n}x,x⟩\right)\right)\right).\hfill \end{array}$$

Hence,

$$f^2\left({\left|⟨\mathbb{A}x,y⟩\right|}_f\right)\le f\left(1\right)f\left(f^{-{\displaystyle \frac{1}{2}}}\left({\displaystyle \sum _{n=1}^{\infty }}{f}^{{\displaystyle \frac{1}{2}}}\left(⟨A_n^{*}{A}_{n}x,x⟩\right)\right)\right).$$

Therefore,

$$f\left({\left|⟨\mathbb{A}x,y⟩\right|}_f\right)\le f\left(1\right)f^{-{\displaystyle \frac{1}{2}}}\left({\displaystyle \sum _{n=1}^{\infty }}{f}^{{\displaystyle \frac{1}{2}}}\left(⟨A_n^{*}{A}_{n}x,x⟩\right)\right)=f\left(1\right){\left|⟨{\mathbb{A}}^{*}\mathbb{A}x,x⟩\right|}_{\sqrt{f}}.$$

(16)

On the other hand, for every $x\in S^1\left(H\right)$ and $0\le \delta \le {\parallel \mathbb{A}\parallel }_f$ it is clear that

$$\begin{array}{cc}\hfill f\left(\delta \right)& \le f(\parallel \mathbb{A}x{\parallel }_f)\hfill \\ \hfill & ={\displaystyle \sum _{n=1}^{\infty }}f\left(\parallel A_{n}x\parallel \right)\hfill \\ \hfill & ={\displaystyle \sum _{n=1}^{\infty }}f\left({⟨A_n^{*}{A}_{n}x,x⟩}^{{\displaystyle \frac{1}{2}}}\right)\hfill \\ \hfill & ={\displaystyle \sum _{n=1}^{\infty }}{f}^{{\displaystyle \frac{1}{2}}}\left(1\right)f^{{\displaystyle \frac{1}{2}}}\left(⟨A_n^{*}{A}_{n}x,x⟩\right)\hfill \\ \hfill & =f^{{\displaystyle \frac{1}{2}}}\left(1\right){\displaystyle \sum _{n=1}^{\infty }}{f}^{{\displaystyle \frac{1}{2}}}\left(\parallel A_n^{*}{A}_{n}x\parallel \right)\hfill \\ \hfill & =f^{{\displaystyle \frac{1}{2}}}\left(1\right)f^{{\displaystyle \frac{1}{2}}}\left(f^{-{\displaystyle \frac{1}{2}}}\left({\displaystyle \sum _{n=1}^{\infty }}{f}^{{\displaystyle \frac{1}{2}}}\left(\parallel A_n^{*}{A}_{n}x\parallel \right)\right)\right)\hfill \\ \hfill & =f^{{\displaystyle \frac{1}{2}}}\left(1\right)f^{{\displaystyle \frac{1}{2}}}\left(\parallel {\mathbb{A}}^{*}{\mathbb{A}x\parallel }_{\sqrt{f}}\right).\hfill \end{array}$$

Thus,

$$f^2\left(\delta \right)\le f\left(1\right)f\left(\parallel {\mathbb{A}}^{*}{\mathbb{A}x\parallel }_{\sqrt{f}}\right).$$

(17)

It follows from (17) it is established that

$$S_{(f,\delta ,q)}^1(\mathbb{A},H)\subset S_{\left(\sqrt{f},f^{-1}\left({\displaystyle \frac{f^2\left(\delta \right)}{f\left(1\right)}}\right)\right)}^1({\mathbb{A}}^{*}\mathbb{A},H).$$

(18)

Therefore, from (16) and (18) it is clear that

$${\omega }_{(f,\delta ,q)}\left(\mathbb{A}\right)\le f\left(1\right){\omega }_{\left(\sqrt{f},f^{-1}\left({\displaystyle \frac{f^2\left(\delta \right)}{f\left(1\right)}}\right)\right)}\left({\mathbb{A}}^{*}\mathbb{A}\right),$$

which completes the proof. □

A relationship between the $(f,\delta ,q)$-numerical radius of an operator and its real and imaginary parts is established in the following result.

Theorem 6. 

Let f be a geometrically convex function. Then, for every operator sequence $\mathbb{A}={\left(A_n\right)}_{n=1}^{\infty }$ with $A_n\in B\left(H\right),n\ge 1,0\le \delta \le \parallel \mathbb{A}\parallel$ and for every $q\in \mathbb{C}$ with $\left|q\right|\le 1$ the inequality

$${\omega }_{(f,\delta ,q)}\left(\mathbb{A}\right)\le {\omega }_{(f,{\delta }_{*},q)}\left(Re\mathbb{A}\right)+{\omega }_{(f,{\delta }_{*},q)}\left(Im\mathbb{A}\right),$$

holds, where ${\delta }_{*}=\delta -\alpha ,\alpha =max\left\{{\parallel Re\mathbb{A}\parallel }_f,{\parallel Im\mathbb{A}\parallel }_f\right\}.$

Proof. 

Since f is strongly increasing and geometrically convex, the midpoint convexity property (see [38]) implies that for every $x,y\in H$ and every $n\ge 1,$

$$f\left(|⟨A_{n}x,y⟩|\right)\le f\left(|⟨Re{A}_{n}x,y⟩|+|⟨Im{A}_{n}x,y⟩|\right).$$

Hence, from [41] it obtained that

$$\begin{array}{cc}\hfill f^{-1}\left({\displaystyle \sum _{n=1}^{\infty }}f\left(|⟨A_{n}x,y⟩|\right)\right)& \le f^{-1}\left({\displaystyle \sum _{n=1}^{\infty }}f\left(|⟨Re{A}_{n}x,y⟩|\right)+{\displaystyle \sum _{n=1}^{\infty }}f\left(|⟨Im{A}_{n}x,y⟩|\right)\right)\hfill \\ \hfill & \le f^{-1}\left({\displaystyle \sum _{n=1}^{\infty }}f\left(|⟨Re{A}_{n}x,y⟩|\right)\right)+f^{-1}\left({\displaystyle \sum _{n=1}^{\infty }}f\left(|⟨Im{A}_{n}x,y⟩|\right)\right).\hfill \end{array}$$

Thus, the last relation can be written in form

$$|⟨\mathbb{A}x,y{|}_f\le |⟨Re\mathbb{A}x,y⟩{|}_f+|⟨Im\mathbb{A}x,y⟩{|}_f,$$

(19)

for any $x,y\in H.$

Next, let $x\in H$ and assume that $0\le \delta \le {\parallel \mathbb{A}\parallel }_f.$ Then, from [41]

$$\begin{array}{cc}\hfill \delta \le {\parallel \mathbb{A}x\parallel }_f& =f^{-1}\left({\displaystyle \sum _{n=1}^{\infty }}f\left(\parallel A_{n}x\parallel \right)\right)\hfill \\ \hfill & \le f^{-1}\left({\displaystyle \sum _{n=1}^{\infty }}f\left(\parallel Re{A}_{n}x\parallel +\parallel Im{A}_{n}x\parallel \right)\right)\hfill \\ \hfill & \le f^{-1}\left({\displaystyle \sum _{n=1}^{\infty }}f\left(\parallel Re{A}_{n}x\parallel \right)\right)+f^{-1}\left({\displaystyle \sum _{n=1}^{\infty }}f\left(\parallel Im{A}_{n}x\parallel \right)\right)\hfill \\ \hfill & \le {\parallel Re\mathbb{A}x\parallel }_f+{\parallel Im\mathbb{A}x\parallel }_f.\hfill \end{array}$$

Let $\alpha =max\left\{{\parallel Re\mathbb{A}\parallel }_f,{\parallel Im\mathbb{A}\parallel }_f\right\}$ and ${\delta }_{*}=\delta -\alpha .$ The preceding inequality implies that ${\delta }_{*}\le {\parallel Re\mathbb{A}x\parallel }_f,{\delta }_{*}\le {\parallel Im\mathbb{A}x\parallel }_f$ and therefore

$$S_{(f,\delta ,q)}^1(\mathbb{A},H)\subset S_{(f,{\delta }_{*},q)}^1(Re\mathbb{A},H)\cap S_{(f,{\delta }_{*},q)}^1(Im\mathbb{A},H).$$

(20)

Hence, from (19), (20) and the definition of the $(f,\delta ,q)$-numerical radius, it is obtained that

$${\omega }_{(f,\delta ,q)}\left(\mathbb{A}\right)\le {\omega }_{(f,{\delta }_{*},q)}\left(Re\mathbb{A}\right)+{\omega }_{(f,{\delta }_{*},q)}\left(Im\mathbb{A}\right).$$

This completes the proof. □

Theorem 7. 

Let $\mathbb{A}={\left(A_n\right)}_{n=1}^{\infty }$ be a sequence of bounded linear operators on $H.$ Then for every $0\le \delta \le {\parallel \mathbb{A}\parallel }_f$ and every $q\in \mathbb{C}$ with $\left|q\right|\le 1,$

$$f\left({\omega }_{(f,\delta ,q)}\left(\mathbb{A}\right)\right)\le {\displaystyle \frac{1}{2}}\left(f\left({\parallel \mathbb{A}\parallel }_f\right)+f\left(\parallel {\mathbb{A}}^{*}{\parallel }_f\right)\right).$$

Proof. 

Let $x,y\in S^1\left(H\right).$ Then

$$f\left({\left|⟨\mathbb{A}x,y⟩\right|}_f\right)={\displaystyle \sum _{n=1}^{\infty }}f\left(|⟨A_{n}x,y⟩|\right)\le {\displaystyle \sum _{n=1}^{\infty }}f\left(\parallel A_{n}x\parallel \right)=f\left({\parallel \mathbb{A}x\parallel }_f\right).$$

(21)

Similarly, for $x,y\in S^1\left(H\right)$

$$f\left({\left|⟨\mathbb{A}x,y⟩\right|}_f\right)={\displaystyle \sum _{n=1}^{\infty }}f\left(|⟨x,A_n^{*}y⟩|\right)\le {\displaystyle \sum _{n=1}^{\infty }}f\left(\parallel A_n^{*}y\parallel \right)=f\left(\parallel {\mathbb{A}}^{*}{y\parallel }_f\right).$$

(22)

Consequently, from (21) and (22)

$$2f\left({\left|⟨\mathbb{A}x,y⟩\right|}_f\right)\le f\left({\parallel \mathbb{A}x\parallel }_f\right)+f\left(\parallel {\mathbb{A}}^{*}{y\parallel }_f\right).$$

Taking supremum over all $(x,y)\in S_{(f,\delta ,q)}^1(\mathbb{A},H),$ and using the definition $(f,\delta ,q)$-numerical radius, we obtain

$$f\left({\omega }_{(f,\delta ,q)}\left(\mathbb{A}\right)\right)\le {\displaystyle \frac{1}{2}}\left(f\left({\parallel \mathbb{A}\parallel }_f\right)+f\left(\parallel {\mathbb{A}}^{*}{\parallel }_f\right)\right).$$

This completes the proof. □

Theorem 8. 

Let f be a geometrically convex function, and let $\mathbb{A}={\left(A_n\right)}_{n=1}^{\infty }$ be a sequence of bounded linear operators on $H.$ Then, for every $0\le \delta \le {\parallel \mathbb{A}\parallel }_f$ and every $q\in \mathbb{C}$ with $\left|q\right|\le 1$

$$f\left({\omega }_{(f,\delta ,q)}\left(\mathbb{A}\right)\right)\le f^{{\displaystyle \frac{1}{2}}}\left(1\right)f^{{\displaystyle \frac{1}{2}}}\left({\omega }_{\left(\sqrt{f},f^{-1}\left({\displaystyle \frac{f\left(\delta \right)}{f^{\frac{1}{2}}\left(1\right)}}\right)\right)}\left({\mathbb{A}}^{*}\mathbb{A}+\mathbb{A}{\mathbb{A}}^{*}\right)\right).$$

Proof. 

Since f is geometrically convex, for every $x,y\in S^1\left(H\right)$ and every $n\ge 1,$ we have

$$\begin{array}{cc}\hfill f\left(|⟨A_{n}x,y⟩|\right)& \le f\left(\parallel A_{n}x\parallel \right)\hfill \\ \hfill & \le f\left(\parallel Re{A}_{n}x\parallel +\parallel Im{A}_{n}x\parallel \right)\hfill \\ \hfill & =f\left(\sqrt{{\left(\parallel Re{A}_{n}x\parallel +\parallel Im{A}_{n}x\parallel \right)}^2}\right)\hfill \\ \hfill & \le f^{{\displaystyle \frac{1}{2}}}\left(1\right)f^{{\displaystyle \frac{1}{2}}}\left({\left(\parallel Re{A}_{n}x\parallel +\parallel Im{A}_{n}x\parallel \right)}^2\right)\hfill \\ \hfill & \le f^{{\displaystyle \frac{1}{2}}}\left(1\right)f^{{\displaystyle \frac{1}{2}}}\left(2\left(\parallel Re{A}_n{x\parallel }^2+{\parallel Im{A}_{n}x\parallel }^2\right)\right)\hfill \\ \hfill & =f^{{\displaystyle \frac{1}{2}}}\left(1\right)f^{{\displaystyle \frac{1}{2}}}\left(⟨2\left(R{e}^2{A}_n+I{m}^2{A}_n\right)x,x⟩\right)\hfill \\ \hfill & =f^{{\displaystyle \frac{1}{2}}}\left(1\right)f^{{\displaystyle \frac{1}{2}}}\left(⟨\left(A_n^{*}{A}_n+A_n{A}_n^{*}\right)x,x⟩\right).\hfill \end{array}$$

Summing over $n\ge 1$ yields

$$f\left({\left|⟨\mathbb{A}x,y⟩\right|}_f\right)\le f^{{\displaystyle \frac{1}{2}}}\left(1\right)f^{{\displaystyle \frac{1}{2}}}\left({⟨\left({\mathbb{A}}^{*}\mathbb{A}+\mathbb{A}{\mathbb{A}}^{*}\right)x,x⟩}_{\sqrt{f}}\right).$$

(23)

Next, using the same technique, it follows that for every $x\in H,$

$$f\left({\parallel \mathbb{A}x\parallel }_f\right)\le f^{{\displaystyle \frac{1}{2}}}\left(1\right)f^{{\displaystyle \frac{1}{2}}}\left(\parallel \left({\mathbb{A}}^{*}\mathbb{A}+\mathbb{A}{\mathbb{A}}^{*}\right){x\parallel }_{\sqrt{f}}\right).$$

Therefore,

$$S_{(f,\delta ,q)}^1\left(\mathbb{A},H\right)\subset S_{\left(\sqrt{f},f^{-1}\left({\displaystyle \frac{f\left(\delta \right)}{f^{\frac{1}{2}}\left(1\right)}}\right)\right)}\left({\mathbb{A}}^{*}\mathbb{A}+\mathbb{A}{\mathbb{A}}^{*},H\right).$$

(24)

Hence, from (23) and (24) we have

$$f\left({\omega }_{(f,\delta ,q)}\left(\mathbb{A}\right)\right)\le f^{{\displaystyle \frac{1}{2}}}\left(1\right)f^{{\displaystyle \frac{1}{2}}}\left({\omega }_{\left(\sqrt{f},f^{-1}\left({\displaystyle \frac{f\left(\delta \right)}{f^{\frac{1}{2}}\left(1\right)}}\right)\right)}\left({\mathbb{A}}^{*}\mathbb{A}+\mathbb{A}{\mathbb{A}}^{*}\right)\right).$$

This completes the proof. □

Theorem 9. 

Let $\mathbb{A}={\left(A_n\right)}_{n=1}^{\infty }$ be an operator sequence on $B\left(H\right),$ where $A_n\in B\left(H\right)$ for $n\ge 1,$ and let f be a geometrically convex function. Then, for every $q\in \mathbb{C}$ with $\left|q\right|\le 1,$ the following inequalities hold:

$${\omega }_{(f,\delta ,q)}\left(Re\mathbb{A}\right)\le {\omega }_{(f,0,q)}\left({\displaystyle \frac{\mathbb{A}}{2}}\right)+{\omega }_{(f,0,\overline{q})}\left({\displaystyle \frac{\mathbb{A}}{2}}\right),$$

for $0\le \delta \le {\parallel Re\mathbb{A}\parallel }_f,$ and

$${\omega }_{(f,\delta ,q)}\left(Im\mathbb{A}\right)\le {\omega }_{(f,0,q)}\left({\displaystyle \frac{\mathbb{A}}{2}}\right)+{\omega }_{(f,0,\overline{q})}\left({\displaystyle \frac{\mathbb{A}}{2}}\right),$$

for $0\le \delta \le {\parallel Im\mathbb{A}\parallel }_f.$

Proof. 

For arbitrary $x,y\in H,$ using the inequality due to [41] it is clear that

$$\begin{array}{cc}\hfill {\left|⟨Re\mathbb{A}x,y⟩\right|}_f& =f^{-1}\left({\displaystyle \sum _{n=1}^{\infty }}f\left(|⟨Re{A}_{n}x,y⟩|\right)\right)\hfill \\ \hfill & \le f^{-1}\left({\displaystyle \sum _{n=1}^{\infty }}f\left({\displaystyle \frac{1}{2}}\left(⟨A_{n}x,y⟩+⟨A_n^{*}x,y⟩\right)\right)\right)\hfill \\ \hfill & =f^{-1}\left({\displaystyle \sum _{n=1}^{\infty }}f\left(\left|{\displaystyle \frac{⟨A_{n}x,y⟩}{2}}\right|+\left|{\displaystyle \frac{⟨A_n^{*}x,y⟩}{2}}\right|\right)\right)\hfill \\ \hfill & =f^{-1}\left({\displaystyle \sum _{n=1}^{\infty }}f\left(\left|⟨{\displaystyle \frac{A_n}{2}}x,y⟩\right|\right)\right)+f^{-1}\left({\displaystyle \sum _{n=1}^{\infty }}f\left(\left|⟨{\displaystyle \frac{A_n^{*}}{2}}x,y⟩\right|\right)\right)\hfill \\ \hfill & =\left({\left|⟨{\displaystyle \frac{\mathbb{A}}{2}}x,y⟩\right|}_f+{\left|\left(⟨{\displaystyle \frac{{\mathbb{A}}^{*}}{2}}x,y\right.⟩\right|}_f\right).\hfill \end{array}$$

(25)

By applying the same argument to the imaginary part, we also have

$${\left|⟨Im\mathbb{A}x,y⟩\right|}_f\le \left({\left|⟨{\displaystyle \frac{\mathbb{A}}{2}}x,y⟩\right|}_f+{\left|⟨{\displaystyle \frac{{\mathbb{A}}^{*}}{2}}x,y⟩\right|}_f\right).$$

(26)

On the other hand, it is easily to see that

$$\begin{array}{cc}\hfill & S_{(f,\delta ,q)}^1(Re\mathbb{A},H)\subset S^1\left(H\right)\times S^1\left(H\right)\mathrm{for}\mathrm{any}0\le \delta \le {\parallel Re\mathbb{A}\parallel }_f,\hfill \\ \hfill & S_{(f,\delta ,q)}^1(Im\mathbb{A},H)\subset S^1\left(H\right)\times S^1\left(H\right)\mathrm{for}\mathrm{any}0\le \delta \le {\parallel Im\mathbb{A}\parallel }_f.\hfill \end{array}$$

(27)

Therefore, from (25), (26) and (27), we have

$${\omega }_{(f,\delta ,q)}\left(Re\mathbb{A}\right)\le {\omega }_{(f,0,q)}\left({\displaystyle \frac{\mathbb{A}}{2}}\right)+{\omega }_{(f,0,q)}\left({\displaystyle \frac{{\mathbb{A}}^{*}}{2}}\right)={\omega }_{(f,0,q)}\left({\displaystyle \frac{\mathbb{A}}{2}}\right)+{\omega }_{(f,0,\overline{q})}\left({\displaystyle \frac{\mathbb{A}}{2}}\right),0\le \delta \le {\parallel Re\mathbb{A}\parallel }_f,$$

and

$${\omega }_{(f,\delta ,q)}\left(Im\mathbb{A}\right)\le {\omega }_{(f,0,q)}\left({\displaystyle \frac{\mathbb{A}}{2}}\right)+{\omega }_{(f,0,q)}\left({\displaystyle \frac{\mathbb{A}}{2}}\right)={\omega }_{(f,0,q)}\left({\displaystyle \frac{\mathbb{A}}{2}}\right)+{\omega }_{(f,0,\overline{q})}\left({\displaystyle \frac{\mathbb{A}}{2}}\right),0\le \delta \le {\parallel Im\mathbb{A}\parallel }_f.$$

This completes the proof. □

Corollary 4. 

If $f\left(t\right)=t^2,0\le t<\infty ,\mathbb{A}=(A_1,\dots ,A_n,0,\dots ),A_m\in B\left(H\right),1\le m\le n$ and $\delta =0,$ then Theorem 9 reduces to the corresponding result obtained in [22].

Theorem 10. 

Let f be a convex function and $\mathbb{A}={\left(A_n\right)}_{n=1}^{\infty }$ be a Hilbert space operator sequence, where $A_n\in B\left(H\right)$ for each $n\ge 1$. If $0\le \delta \le {\parallel \mathbb{A}\parallel }_f$ and $q\in \mathbb{C}$ satisfies $\left|q\right|\le 1$, then

$$f\left({\omega }_{(f,\delta ,q)}^2\left(\mathbb{A}\right)\right)\le {\displaystyle \frac{1}{2}}{\displaystyle \sum _{n=1}^{\infty }}f\left({\left(\left|q\right|+2\sqrt{1-{\left|q\right|}^2}\right)}^2∥A_n^{*}{A}_n+A_n{A}_n^{*}∥\right).$$

Moreover, if f is multiplicative, then

$$f\left({\omega }_{(f,\delta ,q)}^2\left(\mathbb{A}\right)\right)\le {\displaystyle \frac{1}{2}}f\left({\left(\left|q\right|+2\sqrt{1-{\left|q\right|}^2}\right)}^2\right){\displaystyle \sum _{n=1}^{\infty }}f\left(∥A_n^{*}{A}_n+A_n{A}_n^{*}∥\right).$$

Proof. 

For arbitrary $(x,y)\in S_{(f,\delta ,q)}^1(\mathbb{A},H)$, where $0\le \delta \le {\parallel \mathbb{A}\parallel }_f$, $q\in \mathbb{C}$ with $\left|q\right|\le 1$, and for each $n\ge 1$, choose

$$y=\left|q\right|x+\sqrt{1-{\left|q\right|}^2}z,$$

where $\parallel z\parallel =1$ and $⟨z,x⟩=0$.

Moreover, since

$$|⟨A_{n}x,x⟩|\le \omega \left(A_n\right)\mathrm{and}\parallel A_n\parallel \le 2\omega \left(A_n\right),$$

where the latter is the standard inequality relating the operator norm and the numerical radius, by using (1) and (3) a direct computation yields

$$\begin{array}{cc}\hfill f\left(|⟨A_{n}x,y⟩{|}^2\right)& =f\left({\left|⟨A_{n}x,\left|q\right|x+\sqrt{1-{\left|q\right|}^2}z⟩\right|}^2\right)\hfill \\ \hfill & \le f\left({\left(\left|q\right||⟨A_{n}x,x⟩|+\sqrt{1-{\left|q\right|}^2}\left|⟨A_{n}x,z⟩\right|\right)}^2\right)\hfill \\ \hfill & \le f\left({\left(\left|q\right||⟨A_{n}x,x⟩|+\sqrt{1-{\left|q\right|}^2}\parallel A_{n}x\parallel \right)}^2\right)\hfill \\ \hfill & \le f\left({\left(\left|q\right||⟨A_{n}x,x⟩|+\sqrt{1-{\left|q\right|}^2}\parallel A_n\parallel \right)}^2\right)\hfill \\ \hfill & \le f\left({\left(\left|q\right|\omega \left(A_n\right)+2\sqrt{1-{\left|q\right|}^2}\omega \left(A_n\right)\right)}^2\right)\hfill \\ \hfill & =f\left({\left(\left|q\right|+2\sqrt{1-{\left|q\right|}^2}\right)}^2{\omega }^2\left(A_n\right)\right)\hfill \\ \hfill & \le f\left({\displaystyle \frac{1}{2}}{\left(\left|q\right|+2\sqrt{1-{\left|q\right|}^2}\right)}^2∥A_n^{*}{A}_n+A_n{A}_n^{*}∥\right)\hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}f\left({\left(\left|q\right|+2\sqrt{1-{\left|q\right|}^2}\right)}^2∥A_n^{*}{A}_n+A_n{A}_n^{*}∥\right).\hfill \end{array}$$

(28)

Hence, by (28), for every $(x,y)\in S_{f,\delta ,q}(\mathbb{A},H),$ we have

$$f\left({\left|⟨\mathbb{A}x,y⟩\right|}_f^2\right)\le {\displaystyle \frac{1}{2}}{\displaystyle \sum _{n=1}^{\infty }}f\left({\left(\left|q\right|+2\sqrt{1-{\left|q\right|}^2}\right)}^2∥A_n^{*}{A}_n+A_n{A}_n^{*}∥\right).$$

Taking the supremum over all

$$f\left({\omega }_{(f,\delta ,q)}^2\left(\mathbb{A}\right)\right)\le {\displaystyle \frac{1}{2}}{\displaystyle \sum _{n=1}^{\infty }}f\left({\left(\left|q\right|+2\sqrt{1-{\left|q\right|}^2}\right)}^2∥A_n^{*}{A}_n+A_n{A}_n^{*}∥\right).$$

□

Corollary 5. 

As a particular case of Theorem 10, let

$$f\left(t\right)=t,\mathbb{A}=(A,0,\dots ),$$

where $A\in B\left(H\right)$, and let $\delta =0$ and $q=1$. Then

$${\omega }^2\left(A\right)\le {\displaystyle \frac{1}{2}}\parallel A^{*}A+A{A}^{*}\parallel ,$$

which recovers the classical inequality of Kittaneh [5].

Corollary 6. 

As a special case of Theorem 10, let

$$\mathbb{A}=(A,0,\dots ),$$

where $A\in B\left(H\right)$, $\delta =0$, and $q\in \mathbb{C}$ satisfies $\left|q\right|\le 1$. Then Theorem 10 recovers the corresponding result established in [31].

4. Sectorial Case of Coordinate Operators

In this section, the sequential joint $(f,\delta ,q)$-numerical radius function will be investigated in the case when the all coordinate operators are sectorial, as defined in Definition 4.

Theorem 11. 

Let $\mathbb{A}={\left(A_n\right)}_{n=1}^{\infty }$ be an operator sequence on $B\left(H\right),0\le \delta \le {\parallel \mathbb{A}\parallel }_f,$ and $q\in \mathbb{C}$ with $\left|q\right|\le 1.$ Assume that f is convex and multiplicative. Then, the following assertions hold:

(1) 

If $A_n\in Se{c}_{{\alpha }_n}\left(H\right),0\le {\alpha }_n<{\displaystyle \frac{\pi }{2}},n\ge 1$ and ${\alpha }_{sup}=\underset{n\ge 1}{sup}{\alpha }_n<{\displaystyle \frac{\pi }{2}},$ then

$${\omega }_{(f,\delta ,q)}\left(\mathbb{A}\right)\le c_{\alpha }\left({\omega }_{(f,{\delta }_r,q)}\left(Re\mathbb{A}\right)+{\omega }_{(f,{\delta }_i,q)}\left(Im\mathbb{A}\right)\right),$$

where ${\delta }_r={\displaystyle \frac{\delta }{2f^{-1}\left(\frac{1}{2}\right)}}-{\parallel Im\mathbb{A}\parallel }_f,{\delta }_i={\displaystyle \frac{\delta }{2f^{-1}\left(\frac{1}{2}\right)}}-{\parallel Re\mathbb{A}\parallel }_f,$ and$c_{\alpha }=2sec\left({\alpha }_{sup}\right)f^{-1}\left({\displaystyle \frac{1}{2}}\right).$

(2) 

If $A_n\in Se{c}_{{\alpha }_n}\left(H\right),0<{\alpha }_n<{\displaystyle \frac{\pi }{2}},n\ge 1$ and $0<{\alpha }_{inf}=\underset{n\ge 1}{inf}{\alpha }_n<{\displaystyle \frac{\pi }{2}},$ then

$${\omega }_{(f,\delta ,q)}\left(\mathbb{A}\right)\le d_{\alpha }\left({\omega }_{(f,{\delta }_r,q)}\left(Re\mathbb{A}\right)+{\omega }_{(f,{\delta }_i,q)}\left(Im\mathbb{A}\right)\right),$$

where ${\delta }_r={\displaystyle \frac{\delta }{2f^{-1}\left(\frac{1}{2}\right)}}-{\parallel Im\mathbb{A}\parallel }_f,{\delta }_i={\displaystyle \frac{\delta }{2f^{-1}\left(\frac{1}{2}\right)}}-{\parallel Re\mathbb{A}\parallel }_f,$ and$d_{\alpha }=2csc\left({\alpha }_{inf}\right)f^{-1}\left({\displaystyle \frac{1}{2}}\right).$

Proof. (1)

Let $x,y\in H.$ Since $A_n\in Se{c}_{{\alpha }_n}\left(H\right)$ and ${\alpha }_n\le {\alpha }_{sup},$ we have

$$|Im⟨A_{n}x,y⟩|\le \tan \left({\alpha }_{sup}\right)\left|Re⟨A_{n}x,y⟩\right|,$$

for every $n\ge 1.$ Hence, since f is multiplicative, then for every $n\ge 1$

$$\begin{array}{cc}\hfill f\left(|⟨A_{n}x,y⟩|\right)& =f\left({\left(|Re⟨A_{n}x,y⟩{|}^2+{\left|Im⟨A_{n}x,y⟩\right|}^2\right)}^{{\displaystyle \frac{1}{2}}}\right)\hfill \\ \hfill & \le f\left({\left(|Re⟨A_{n}x,y⟩{|}^2+{\tan }^2\left({\alpha }_{sup}\right){\left|Re⟨A_{n}x,y⟩\right|}^2\right)}^{{\displaystyle \frac{1}{2}}}\right)\hfill \\ \hfill & =f\left(sec\left({\alpha }_{sup}\right)\left|Re⟨A_{n}x,y⟩\right|\right)\hfill \\ \hfill & =f(sec\left({\alpha }_{sup}\right))f\left(|Re⟨A_{n}x,y⟩|\right).\hfill \end{array}$$

(29)

On the other hand, we get for every $x,y\in H$

$$\begin{array}{cc}\hfill ⟨A_{n}x,y⟩& =⟨(Re{A}_n+iIm{A}_n)x,y⟩\hfill \\ \hfill & =⟨Re{A}_{n}x,y⟩+i⟨Im{A}_{n}x,y⟩\hfill \\ \hfill & =Re⟨Re{A}_{n}x,y⟩+iIm⟨Re{A}_{n}x,y⟩+iRe⟨Im{A}_{n}x,y⟩-Im⟨Im{A}_{n}x,y⟩\hfill \\ \hfill & =\left(Re⟨Re{A}_{n}x,y⟩-Im⟨Im{A}_{n}x,y⟩\right)+i\left(Im⟨Re{A}_{n}x,y⟩+Re⟨Im{A}_{n}x,y⟩\right)\hfill \\ \hfill & =Re⟨A_{n}x,y⟩+iIm⟨A_{n}x,y⟩,\hfill \end{array}$$

where

$$Re⟨A_{n}x,y⟩=Re⟨Re{A}_{n}x,y⟩-Im⟨Im{A}_{n}x,y⟩,$$

and

$$Im⟨A_{n}x,y⟩=Im⟨Re{A}_{n}x,y⟩+Re⟨Im{A}_{n}x,y⟩.$$

Then, from (29), the last relations and midpoint convexity property of f

$$\begin{array}{cc}\hfill f\left(|⟨A_{n}x,y⟩|\right)& \le f(sec\left({\alpha }_{sup}\right))f\left(|Re⟨Re{A}_{n}x,y⟩-Im⟨Im{A}_{n}x,y⟩|\right)\hfill \\ \hfill & \le f(sec\left({\alpha }_{sup}\right))f\left(|Re⟨Re{A}_{n}x,y⟩|+|Im⟨Im{A}_{n}x,y⟩|\right)\hfill \\ \hfill & =f(sec\left({\alpha }_{sup}\right))f\left(2{\displaystyle \frac{|Re⟨Re{A}_{n}x,y⟩|+|Im⟨Im{A}_{n}x,y⟩|}{2}}\right)\hfill \\ \hfill & ={\displaystyle \frac{f(2sec\left({\alpha }_{sup}\right))}{2}}\left(f\left(|Re⟨Re{A}_{n}x,y⟩|\right)+f\left(|Im⟨Im{A}_{n}x,y⟩|\right)\right)\hfill \\ \hfill & \le {\displaystyle \frac{f(2sec\left({\alpha }_{sup}\right))}{2}}\left(f\left(|⟨Re{A}_{n}x,y⟩|\right)+f\left(|⟨Im{A}_{n}x,y⟩|\right)\right).\hfill \end{array}$$

Summing over n and applying $f^{-1},$ it follows that

$$\begin{array}{cc}\hfill f^{-1}\left({\displaystyle \sum _{n=1}^{\infty }}f\left(|⟨A_{n}x,y⟩|\right)\right)& \le 2sec\left({\alpha }_{sup}\right)f^{-1}\left({\displaystyle \frac{1}{2}}\right)\left(f^{-1}\left({\displaystyle \sum _{n=1}^{\infty }}f\left(|⟨Re{A}_{n}x,y⟩|\right)\right)\right.\hfill \\ \hfill & \left.+f^{-1}\left({\displaystyle \sum _{n=1}^{\infty }}f\left(|⟨Im{A}_{n}x,y⟩|\right)\right)\right).\hfill \end{array}$$

Consequently, for $x,y\in H$

$${\left|⟨\mathbb{A}x,y⟩\right|}_f\le c_{\alpha }\left({\left|⟨Re\mathbb{A}x,y⟩\right|}_f+{\left|⟨Im\mathbb{A}x,y⟩\right|}_f\right),$$

(30)

where $c_{\alpha }=2sec\left({\alpha }_{sup}\right)f^{-1}\left({\displaystyle \frac{1}{2}}\right).$

Now, let $0\le \delta \le {\parallel \mathbb{A}\parallel }_f$ and $x\in S^1\left(H\right).$ Since f is multiplicative, by the midpoint property of f, we have

$$\begin{array}{cc}\hfill \delta & \le {\parallel \mathbb{A}x\parallel }_f\hfill \\ \hfill & =f^{-1}\left({\displaystyle \sum _{n=1}^{\infty }}f\left(\parallel A_{n}x\parallel \right)\right)\hfill \\ \hfill & \le f^{-1}\left({\displaystyle \sum _{n=1}^{\infty }}f\left(\parallel Re{A}_{n}x\parallel +\parallel Im{A}_{n}x\parallel \right)\right)\hfill \\ \hfill & =f^{-1}\left({\displaystyle \sum _{n=1}^{\infty }}f\left(2{\displaystyle \frac{\parallel Re{A}_{n}x\parallel +\parallel Im{A}_{n}x\parallel }{2}}\right)\right)\hfill \\ \hfill & \le f^{-1}\left({\displaystyle \sum _{n=1}^{\infty }}f\left(2\right){\displaystyle \frac{f\left(\parallel Re{A}_{n}x\parallel \right)+f\left(\parallel Im{A}_{n}x\parallel \right)}{2}}\right)\hfill \\ \hfill & =2f^{-1}\left({\displaystyle \frac{1}{2}}\right)\left(f^{-1}\left({\displaystyle \sum _{n=1}^{\infty }}f\left(\parallel Re{A}_{n}x\parallel \right)\right)+f^{-1}\left({\displaystyle \sum _{n=1}^{\infty }}f\left(\parallel Im{A}_{n}x\parallel \right)\right)\right)\hfill \\ \hfill & \le 2f^{-1}\left({\displaystyle \frac{1}{2}}\right)\left({\parallel Re\mathbb{A}x\parallel }_f+{\parallel Im\mathbb{A}x\parallel }_f\right).\hfill \end{array}$$

Then, it is clear that

$${\delta }_r\le {\parallel Re\mathbb{A}x\parallel }_f,x\in S^1\left(H\right),$$

where ${\delta }_r={\displaystyle \frac{\delta }{2f^{-1}\left({\displaystyle \frac{1}{2}}\right)}}-{\parallel Im\mathbb{A}\parallel }_f$ and similarly it can be shown that if $y\in S^1\left(H\right)$ and $\delta \le \parallel {\mathbb{A}}^{*}{y\parallel }_f,$ then ${\delta }_i\le {\parallel Re\mathbb{A}y\parallel }_f,$ where ${\delta }_i={\displaystyle \frac{\delta }{2f^{-1}\left({\displaystyle \frac{1}{2}}\right)}}-{\parallel Re\mathbb{A}\parallel }_f.$ Therefore,

$$S_{(f,\delta ,q)}^1(\mathbb{A},H)\subset S_{(f,{\delta }_r,q)}^1(Re\mathbb{A},H)$$

and

$$S_{(f,\delta ,q)}^1(\mathbb{A},H)\subset S_{(f,{\delta }_i,q)}^1(Im\mathbb{A},H).$$

Then, from above calculations it is obtained that

$$S_{(f,\delta ,q)}^1(\mathbb{A},H)\subset S_{(f,{\delta }_r,q)}^1(Re\mathbb{A},H)\cap S_{(f,{\delta }_i,q)}^1(Im\mathbb{A},H).$$

(31)

Hence, from (30) and (31) it is established that

$${\omega }_{(f,\delta ,q)}\left(\mathbb{A}\right)\le c_{\alpha }\left({\omega }_{(f,{\delta }_r,q)}\left(Re\mathbb{A}\right)+{\omega }_{(f,{\delta }_i,q)}\left(Im\mathbb{A}\right)\right),$$

This proves the first assertion.

(2)

The proof of the second assertion follows analogously.

Thus, the theorem is completely proved.

□

Theorem 12. 

Let $q=1$ and let $\mathbb{A}={\left(A_n\right)}_{n=1}^{\infty }$ be an operator sequences on $B\left(H\right).$ Assume that for every $n\ge 1$ the coordinate operators $A_n$ satisfy

$$W_1\left(A_n\right)\subset \left\{r{e}^{i\phi }:r\ge 0,0\le {\gamma }_n\le \phi \le {\alpha }_n<{\displaystyle \frac{\pi }{2}}\right\}.$$

Then,

(1) 

For any $0\le \delta \le \parallel sec\left({\gamma }_{inf}\right){Re\mathbb{A}\parallel }_f,$

$${\omega }_{(f,\delta ,1)}(sec\left({\gamma }_{inf}\right)Re\mathbb{A})\le {\omega }_{(f,0,1)}\left(\mathbb{A}\right),$$

where $0\le {\gamma }_{inf}=\underset{n\ge 1}{inf}{\gamma }_n<{\displaystyle \frac{\pi }{2}}.$

(2) 

For any $0\le \delta \le \parallel csc\left({\alpha }_{sup}\right){Im\mathbb{A}\parallel }_f,$

$${\omega }_{(f,\delta ,1)}(csc\left({\alpha }_{sup}\right)Im\mathbb{A})\le {\omega }_{(f,0,1)}\left(\mathbb{A}\right),$$

where $0<{\alpha }_{sup}=\underset{n\ge 1}{sup}{\alpha }_n<{\displaystyle \frac{\pi }{2}}.$

(3) 

For any $0\le \delta \le {\parallel \mathbb{A}\parallel }_f,$

$${\omega }_{(f,\delta ,1)}\left(\mathbb{A}\right)\le {\omega }_{(f,0,1)}(sec\left({\alpha }_{sup}\right)Re\mathbb{A}),$$

where $0\le {\alpha }_{sup}=\underset{n\ge 1}{sup}{\alpha }_n<{\displaystyle \frac{\pi }{2}}.$

(4) 

For any $0\le \delta \le {\parallel \mathbb{A}\parallel }_f,$

$${\omega }_{(f,\delta ,1)}\left(\mathbb{A}\right)\le {\omega }_{(f,0,1)}(csc\left({\alpha }_{inf}\right)Im\mathbb{A}),$$

where $0<{\alpha }_{inf}=\underset{n\ge 1}{inf}{\alpha }_n<{\displaystyle \frac{\pi }{2}}.$

Proof. 

Let $x\in H$ be arbitrary. Then,

$$f\left({\left|⟨\mathbb{A}x,x⟩\right|}_f\right)={\displaystyle \sum _{n=1}^{\infty }}f\left(|⟨A_{n}x,x⟩{|}_f\right)={\displaystyle \sum _{n=1}^{\infty }}f\left({\left(|⟨Re{A}_{n}x,x⟩{|}^2+{\left|⟨Im{A}_{n}x,x⟩\right|}^2\right)}^{{\displaystyle \frac{1}{2}}}\right)$$

On the other hand, since $x\in H$

$$|⟨Im{A}_{n}x,x⟩|\ge \tan \left({\gamma }_n\right)\left|⟨Re{A}_{n}x,x⟩\right|,n\ge 1$$

and the tangent function is monotonically increasing in $\left[0,{\displaystyle \frac{\pi }{2}}\right),$ it follows that

$$\begin{array}{cc}\hfill f\left({\left|⟨\mathbb{A}x,x⟩\right|}_f\right)& \ge {\displaystyle \sum _{n=1}^{\infty }}f\left({\left(1+{\tan }^2\left({\gamma }_n\right)\right)}^{{\displaystyle \frac{1}{2}}}\left|⟨Re\mathbb{A}x,x⟩\right|\right)\hfill \\ \hfill & \ge {\displaystyle \sum _{n=1}^{\infty }}f\left({\left(1+{\tan }^2\left({\gamma }_{inf}\right)\right)}^{{\displaystyle \frac{1}{2}}}\left|⟨Re\mathbb{A}x,x⟩\right|\right)\hfill \\ \hfill & =f\left(|⟨sec\left({\gamma }_{inf}\right)Re\mathbb{A}x,x⟩|\right).\hfill \end{array}$$

Thus, for any $x\in H,$

$${\left|⟨\mathbb{A}x,x⟩\right|}_f\ge {\left|⟨sec\left({\gamma }_{inf}\right)Re\mathbb{A}x,x⟩\right|}_f.$$

(32)

On the other hand, for $0\le \delta \le \parallel sec\left({\gamma }_{inf}\right){Re\mathbb{A}x\parallel }_f,$ the following inclusion holds

$$S_{(f,\delta ,1)}^1\left(sec\left({\gamma }_{inf}\right)Re\mathbb{A},H\right)\subset S^1\left(H\right).$$

(33)

Then, from (32) and (33) it is obtained that

$${\omega }_{(f,\delta ,1)}(sec\left({\gamma }_{inf}\right)Re\mathbb{A})\le {\omega }_{(f,0,1)}\left(\mathbb{A}\right).$$

This proves the first assertion.

(2)–(4) The proofs are analogous to that of assertion (1).

Hence, the theorem is completely proved. □

Corollary 7. 

In case when $\mathbb{A}=(A,0,\dots ),A\in B\left(H\right),\delta =0,$ and $q=1,$ the result in assertion (2) of Theorem 12 coincides with claim obtained in [46].

5. Conclusions

In this paper, we introduced the sequential joint $(f,\delta ,q)$-numerical radius for sequences of bounded linear operators on Hilbert spaces and developed its basic theory. We established fundamental properties of this generalized numerical radius, including its behaviour under scalar multiplication, adjoint operations, and decompositions involving the real and imaginary parts of operator sequences.

Furthermore, we obtained several upper and lower estimates for the sequential joint $(f,\delta ,q)$-numerical radius. These inequalities provide a unified framework that extends a number of existing numerical radius inequalities and recovers several known results as special cases. We also investigated the interaction between the introduced numerical radius and sectorial operator sequences, showing how geometric restrictions on the numerical ranges of the coordinate operators lead to additional bounds and structural relations.

The results presented here demonstrate that the sequential joint $(f,\delta ,q)$-numerical radius provides a flexible framework for studying operator sequences and their associated inequalities. The introduced concept connects generalized numerical radius theory with geometric properties of operator ranges and offers new perspectives for the analysis of sequences of Hilbert space operators.

Several directions remain open for future investigation. In particular, it would be interesting to study analogous constructions for broader classes of operators, including unbounded operators, operators on more general Banach spaces, and other generalized numerical radius functions. Further questions concerning applications to operator means, matrix inequalities, and related areas of functional analysis may also provide fruitful directions for future research.